Geometrical Optics | Physics JEE previous year questions with Solutions

Q 1 :

An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is 25 D. Focal length of each of the convex lens is [2024]

20 cm
50 cm
25 cm
500 cm

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(1)

We know that $P_{e q} = Σ P_{i}$

As given all lenses are identical, 5P = 25 D

$∴ P = 5 D$

$\Rightarrow \frac{1}{f} = 5 \Rightarrow f = \frac{1}{5} m = 20 c m$

Q 2 :

Given below are two statements:

Statement (I): When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.

Statement (II): Concave lens always forms a virtual and erect image.

In the light of the above statements, choose the correct answer from the options given below: [2024]

Statement I is false but Statement II is true.
Both Statement I and Statement II are false.
Statement I is true but Statement II is false.
Both Statement I and Statement II are true.

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(1)

Statement-I is False. (Image is formed on the same side, not on the other side).

Statement-II is True. (As object is real, its virtual and erect image is formed).

Q 3 :

A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be: [2024]

-16 cm
-160 cm
+160 cm
+16 cm

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(2)

$μ_{0} = 1.5, μ_{liq} = 1.6 a n d f_{a i r} = 20 c m$

$A s \frac{f_{l i q}}{f_{a i r}} = \frac{(μ_{0} - 1) μ_{l i q}}{(μ_{0} - μ_{l i q})} \Rightarrow \frac{f_{l i q}}{20} = \frac{(1.5 - 1) 1.6}{(1.5 - 1.6)}$

$\Rightarrow f_{l i q} = - 160 c m$

Q 4 :

For the thin convex lens, the radii of curvature are 15 cm and 30 cm respectively. The focal length of the lens is 20 cm. The refractive index of the material is [2024]

1.4
1.2
1.5
1.8

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(3)

$\frac{1}{f} = (\frac{μ_{lens}}{μ_{air}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\Rightarrow \frac{1}{20} = (\frac{μ}{1} - 1) (\frac{1}{15} - \frac{1}{(- 30)})$

$\Rightarrow \frac{1}{20} = (μ - 1) (\frac{3}{30})$

$\Rightarrow μ - 1 = \frac{1}{2} \Rightarrow μ = 1 + \frac{1}{2} = \frac{3}{2} = 1.5$

Q 5 :

The position of the image formed by the combination of lenses is [2024]

30 cm (left of third lens)
15 cm (right of second lens)
15 cm (left of second lens)
30 cm (right of third lens)

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(4)

Here we will use the lens formula three times, for lens 1

$u = - 30 cm, f = 10 cm$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v_{1}} - \frac{1}{- 30} = \frac{1}{10} \Rightarrow v_{1} = 15 cm$

For 2nd lens $u = 10 cm, f = - 10 cm$

$v_{2} \to \infty$

For 3rd lens: $f = 30 cm, u = - \infty, v = ?$

So $v$ will be 30 cm.

Q 6 :

The following figure represents two biconvex lenses $L_{1}$ and $L_{2}$ having focal length 10 cm and 15 cm respectively. The distance between $L_{1}$ and $L_{2}$ is [2024]

10 cm
15 cm
25 cm
35 cm

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(3)

$D = f_{1} + f_{2} = 25$ cm, Paraxial parallel rays pass through focus and ray from focus of convex lens will become parallel.

Q 7 :

The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is __________ cm. [2024]

(10)

$3 x = 45 cm$ $x = 15 cm$

$u = - 15 cm$ $v = + 30 cm$

By lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{30} - \frac{1}{- 15} = \frac{1}{f}$

$\frac{1 + 2}{30} = \frac{1}{f}$

$f = 10 cm$

Q 8 :

In an experiment to measure the focal length ( $f$ ) of a convex lens, the magnitude of object distance ( $x$ ) and the image distance ( $y$ ) are measured with reference to the focal point of the lens. The $y - x$ plot is shown in figure.

The focal length of the lens is _____ cm. [2024]

(20)

Here, $u = 20 cm$

From the focal point, $v = 20 cm$

$x y = f^{2} \to$ Newton's formula

$20 \times 20 = f^{2}$

$f = 20$

Q 9 :

Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at _____ cm from the object. [2024]

(200)

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.5}{v} - \frac{1}{- 100} = \frac{1.5 - 1}{20}$

$v = 100 cm$

Distance from object $= 100 + 100 = 200 cm$

Q 10 :

The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is ______ cm. [2024]

(15)

$m = \frac{v}{u} \Rightarrow v = 3 u$

$∵$ $v - u = 20$

$3 u - u = 20 \Rightarrow u = 10 c m,$ $v = 30 c m$

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = - \frac{1}{30} + \frac{1}{10} = \frac{- 1 + 3}{30} = \frac{2}{30} \Rightarrow f = 15 cm$

Q 11 :

In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to [ $R_{1}$ ] and [ $R_{2}$ ], i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is [2025]

$- \frac{1}{6} (\frac{1}{| R_{1} |} + \frac{1}{| R_{2} |})$
$- \frac{1}{6} (\frac{1}{| R_{1} |} - \frac{1}{| R_{2} |})$
$\frac{1}{6} (\frac{1}{| R_{1} |} + \frac{1}{| R_{2} |})$
$\frac{1}{6} (\frac{1}{| R_{1} |} - \frac{1}{| R_{2} |})$

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(2)

$\Rightarrow P_{eq} = P_{1} + P_{2} + P_{3}$

$\Rightarrow P_{1} = (\frac{4}{3} - 1) (\frac{1}{\infty} - \frac{1}{- | R_{1} |})$

$\Rightarrow P_{1} = (\frac{1}{3 | R_{1} |})$

$\Rightarrow P_{2} = (\frac{1}{2}) (\frac{1}{- | R_{1} |} - \frac{1}{- | R_{2} |})$

$\Rightarrow P_{2} = \frac{1}{2} (\frac{1}{| R_{2} |} - \frac{1}{| R_{1} |})$

$\Rightarrow P_{3} = (\frac{1}{3}) (\frac{1}{- | R_{2} |} - \frac{1}{\infty}) = - \frac{1}{3 | R_{2} |}$

$\Rightarrow P_{e q} = \frac{1}{3} (\frac{1}{| R_{1} |} - \frac{1}{| R_{2} |}) - \frac{1}{2} (\frac{1}{| R_{1} |} - \frac{1}{| R_{2} |})$

$= - \frac{1}{6} (\frac{1}{| R_{1} |} - \frac{1}{| R_{2} |})$

Q 12 :

Given is a thin convex lens of glass (refractive index $μ$ ) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself. [2025]

$R / μ$
$R / (2 μ - 3)$
$μ R$
$R / (2 μ - 1)$

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(4)

$P_{eq} = 2 P_{l} + P_{m} \Rightarrow - \frac{1}{f_{eq}} = \frac{2}{f_{l}} - \frac{1}{f_{m}}$

$- \frac{1}{f_{eq}} = \frac{4 (μ - 1)}{R} - \frac{2}{- R} = \frac{1}{R} (4 μ - 4 + 2)$

$- \frac{1}{f_{eq}} = \frac{1}{R} (4 μ - 2) \Rightarrow \frac{1}{f_{eq}} = \frac{- 1}{R} (4 μ - 2)$

$f_{eq} = \frac{R}{2} \Rightarrow R = 2 f_{eq} = - 2 (\frac{R}{4 μ - 2}) = \frac{- R}{(2 μ - 1)}$

For concave mirror, object should be at 2f for the image to be at same point

Distance = $\frac{R}{(2 μ - 1)}$

Q 13 :

A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4D then the power of a part of the divided lens is [2025]

8D
4D
1D
2D

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(4)

$P = \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$4 D = (μ - 1) \frac{2}{R}$ ...(i)

$P = (μ - 1) (\frac{1}{R}) = 2 D$

Q 14 :

A spherical surface of radius of curvature R, separates air from glass (refractive index = 1.5). The centre of curvature is in the glass medium. A point object 'O' placed in air on the optic axis of the surface, so that its real image is formed at ' $I$ ' inside glass. The line O $I$ intersects the spherical surface at P and PO = P $I$ . The distance PO equals to: [2025]

5 R
3 R
2 R
1.5 R

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(1)

PO = u = –x

P $I$ = v = x

PO = P $I$

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.5}{x} + \frac{1}{x} = \frac{1}{2 R}$

$\frac{5}{2 x} = \frac{1}{2 R} \Rightarrow x = 5 R$

Q 15 :

Given a thin convex lens (refractive index $μ_{2}$ ), kept in a liquid (refractive index $μ_{1}, μ_{1} < μ_{2}$ ) having radii of curvature | $R_{1}$ | and | $R_{2}$ |. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place? [2025]

$\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (| R_{1} | + | R_{2} |) - μ_{1} | R_{1} |}$
$\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (| R_{1} | + | R_{2} |) - μ_{1} | R_{2} |}$
$\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (2 | R_{1} | + | R_{2} |) - μ_{1} \sqrt{| R_{1} | \cdot | R_{2} |}}$
$\frac{(μ_{2} + μ_{1}) | R_{1} |}{(μ_{2} - μ_{1})}$

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(2)

$\frac{1}{f_{l}} = (\frac{μ_{2}}{μ_{1}} - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$f_{m} = - \frac{R_{2}}{2}$

$P_{eq} = 2 P_{l} + P_{m}$

$\frac{1}{| f |} = \frac{2 (μ_{2} - μ_{1})}{μ_{1}} (\frac{1}{R_{1}} + \frac{1}{R_{2}}) + \frac{2}{R_{2}}$

$= \frac{2 (μ_{2} - μ_{1}) (R_{1} + R_{2})}{μ_{1} (R_{1} R_{2})} + \frac{2}{R_{2}}$

$\frac{1}{f} = \frac{2 μ_{2} R_{1} + 2 μ_{2} R_{2} - 2 μ_{1} R_{1} - 2 μ_{1} R_{2} + 2 μ_{1} R_{1}}{μ_{1} R_{1} R_{2}}$

$\Rightarrow f = \frac{μ_{1} R_{1} R_{2}}{2 μ_{2} R_{1} + 2 μ_{2} R_{2} - 2 μ_{1} R_{2}}$

Required distance = $2 f$

Q 16 :

What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D? ['D' stands for dioptre] [2025]

0.04
0.40
0.1
0.01

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(1)

When P = 2.5 D

$F = \frac{1}{P} = \frac{1}{2.5}$

When $P' = 2.6 D$

$F' = \frac{1}{P'} = \frac{1}{2.6}$

Relative decrease in focal length

$\frac{F - F'}{F} = \frac{\frac{2}{5} - \frac{5}{13}}{\frac{2}{5}} = 1 - \frac{25}{26} = \frac{1}{26} = 0.04$

Q 17 :

A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is [2025]

0.15 m
0.10 m
0.20 m
0.25 m

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(2)

The system act as a concave mirror of focal length 0.2 metre, it means parallel rays of light gets focused at a distance 0.2 metre in front of the polished lens.

Using, $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{(μ_{2} - μ_{1})}{R}$

Here, $μ_{1} = 1.2, μ_{2} = 1.5 and u = \infty$

$\frac{1.5}{v} - \frac{1.2}{\infty} = \frac{(1.5 - 1.2)}{R} \Rightarrow v = 5 R$

The image formed by the curved surface acts as virtual object for plane mirror and it will from a real image in front of the plane mirror. Now rays again interact with curved surface and form final image.

Again using, $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{(μ_{2} - μ_{1})}{R}$

Here, $μ_{1} = 1.5, μ_{2} = 1.2, u = 5 R and v = 0.2$

$\frac{1.2}{0.2} - \frac{1.5}{5 R} = \frac{(1.2 - 1.5)}{- R} \Rightarrow R = \frac{1}{10} m = 10 cm$

Q 18 :

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of $f_{1}$ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of $f_{2}$ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of $f_{1}$ and $f_{2}$ will be: [2025]

3 : 5
1 : 3
1 : 2
2 : 3

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(2)

$\frac{1}{f_{1}} = (1.5 - 1) [\frac{1}{2} - 0] \Rightarrow f_{1} = 4 cm$

$\frac{1}{f_{2}} = (\frac{1.5}{1.2} - 1) (\frac{1}{3} - 0)$

$\frac{1}{f_{2}} = \frac{0.3}{1.2} \times \frac{1}{3} \Rightarrow f_{2} = 12 cm$

$\Rightarrow \frac{f_{1}}{f_{2}} = 4 : 12 = 1 : 3$

Q 19 :

A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is 2 cm $\times$ 2 cm and the area of the landscape photographed is 400 ${km}^{2}$ . The focal length of the lens in the drone camera is [2025]

1.8 cm
2.8 cm
2.5 cm
0.9 cm

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(1)

Linear size of image = 2 cm

Linear size of object $= 20 \times 10^{5}$ cm

$m = \frac{2}{20 \times 10^{5}} = 10^{- 6}$

So, $v = \frac{u}{10^{6}}$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{10^{6}}{u} + \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{(1 + 10^{6})}{u} = \frac{1}{f} \Rightarrow f \approx (\frac{u}{10^{6}})$

$\Rightarrow f = \frac{18000 \times 100}{10^{6}} cm = 1.8 cm$

Q 20 :

In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of '–2' then the radius of curvature of meniscus is: [2025]

1 cm
$\frac{1}{3} cm$
$\frac{2}{3} cm$
$\frac{4}{3} cm$

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(3)

$μ_{A} = 1.3, μ_{B} = 1.4 and u = - 13 cm$

$m = \frac{\frac{v}{μ_{B}}}{\frac{u}{μ_{A}}} = - 2$

$\Rightarrow v = - 2 u (\frac{μ_{B}}{μ_{A}})$

$= - 2 (- 13 cm) (\frac{1.4}{1.3}) = + 28 cm$

Also, $\frac{μ_{B}}{v} - \frac{μ_{A}}{u} = \frac{μ_{B} - μ_{A}}{R}$

$\frac{1.4}{28} - \frac{1.3}{- 13} = \frac{0.1}{R} \Rightarrow R = \frac{2}{3} cm$

Q 21 :

Two identical symmetric double convex lenses of focal length f are cut into two equal parts $L_{1}, L_{2}$ by AB plane and $L_{3}, L_{4}$ by XY plane as shown in figure respectively. The ratio of focal length of lenses $L_{1}$ and $L_{3}$ is [2025]

1 : 4
1 : 1
2 : 1
1 : 2

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(4)

So, $\frac{f_{1}}{f_{3}} = 1 : 2$

Q 22 :

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is: [2025]

0.214 R
0.114 R
0.411 R
0.124 R

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(2)

For B

$\frac{μ_{2}}{V_{B}} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.5}{V_{B}} + \frac{2}{R} = \frac{0.5}{- R}$

$\frac{1.5}{V_{B}} = - \frac{1}{2 R} - \frac{2}{R}$

$\frac{1.5}{V_{B}} = \frac{- 5}{2 R} \Rightarrow V_{B} = - 0.6 R$

For A

$\frac{1.5}{V_{A}} + \frac{2}{3 R} = \frac{0.5}{- R}$

$\frac{1.5}{V_{A}} = - \frac{1}{2 R} - \frac{2}{3 R}$

$\frac{1.5}{V_{A}} = - \frac{7}{6 R} \Rightarrow V_{A} = - \frac{9}{7} R$

Distance between images $= 2 R - (0.6 R + \frac{9}{7} R) = 0.114 R$

Q 23 :

A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to [2025]

72 cm
96 cm
24 cm
48 cm

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(2)

$\frac{1}{24} = (\frac{μ_{l}}{μ_{s}} - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{24} = (1.5 - 1) [\frac{2}{R}]$ ... (i)

$\frac{1}{f'} = (\frac{1.5}{1.33} - 1) (\frac{2}{R})$ ... (ii)

(i) divided by (ii), $\frac{f'}{24} = 4 \Rightarrow f' = 96 cm$

Q 24 :

A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is: [2025]

$- \frac{α}{2}$
–45°
+45°
$- α$

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(2)

Location of image of A :

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{- 30} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{60} \Rightarrow v = 60 cm$

$m = \frac{v}{u} = \frac{60}{- 30} = - 2$

Since size of object is small w.r.t. the location hence

$d v = m^{2} d u \Rightarrow d v = 4 \times 1 = 4 cm$

$d y = h_{i} = m h_{0} \Rightarrow h_{i} = 2 \times 2 = 4 cm$

$∴$ Angle made with principle axis = –45°

Q 25 :

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

0.24 m right to the spherical surface
0.4 m left to the spherical surface
0.24 m left to the spherical surface
0.4 m right to the spherical surface

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(2)

For spherical surfaces

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1.5}{v} - \frac{1}{(- 0.2)} = \frac{1.5 - 1}{0.4}$

$\frac{1.5}{v} = \frac{0.5}{0.4} - \frac{1}{0.2}$

$\Rightarrow v = - 0.4 m$

Q 26 :

A bi-convex lens has radius of curvature of both the surfaces same as 1/6 cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides ( $R_{1} \neq R_{2}$ ), without any change in lens power then possible combination of $R_{1}$ and $R_{2}$ is : [2025]

$\frac{1}{3} cm and \frac{1}{3} cm$
$\frac{1}{5} cm and \frac{1}{7} cm$
$\frac{1}{3} cm and \frac{1}{7} cm$
$\frac{1}{6} cm and \frac{1}{9} cm$

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(2)

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

In 1st case:

$\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R})$

$= (μ - 1) (\frac{2}{R}) = (μ - 1) 12$

in 2nd case:

$\frac{1}{f'} = (μ - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

If $f' = f \Rightarrow 12 = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

Hence, option (2) is correct.

Q 27 :

The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm. The refractive index of the lens material is [2025]

1.2
1.4
1.5
1.8

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(3)

For bi-convex

$R_{1} = 10 cm, R_{2} = - 15 cm$

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{12} = (μ - 1) (\frac{1}{10} - \frac{1}{- 15}) = (μ - 1) (\frac{1}{6})$

$μ - 1 = \frac{1}{2} \Rightarrow μ = 1.5$

Q 28 :

A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when it is placed in water.

(Take refractive index of water as 1.28) [2025]

355 mm
288 mm
555 mm
655 mm

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(2)

As we know,

$\frac{1}{f} = [\frac{μ_{L}}{μ_{m}} -] [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

For air $μ_{m} = 1$

$\frac{1}{12} = [1.6 - 1] [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{12} = \frac{6}{10} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$[\frac{1}{R_{1}} - \frac{1}{R_{2}}] = \frac{10}{72}$ ...(i)

For water, $μ_{m} = 1.28$

$\frac{1}{f} = [\frac{1.6}{1.28} - 1] [\frac{10}{72}] = \frac{32}{128} \times \frac{10}{72}$

$\frac{1}{f} = \frac{1}{4} \times \frac{10}{72} \Rightarrow f$ =28.8 cm = 288 mm

Q 29 :

Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is: [2025]

5 D
1 D
20 D
10 D

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(4)

$f_{1} = 30 cm, f_{2} = 10 cm$

$\frac{1}{f_{eq}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}}$ , d = distance between lens

$\frac{1}{f_{eq}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{(0.3) (0.1)}$

$\frac{1}{f_{eq}} = \frac{1}{0.1}$

Power = $\frac{1}{f_{eq}} = 10 D$

Q 30 :

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be [2025]

$\frac{500}{11} cm$
$\frac{800}{11} cm$
$\frac{700}{11} cm$
$\frac{600}{11} cm$

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(4)

$\frac{1}{f} = (1.3 - 1) (\frac{1}{\infty} - \frac{1}{- 30}) + (1.5 - 1) (\frac{1}{- 30} - \frac{1}{- 20})$

$= \frac{0.3}{30} + \frac{0.5}{60} = \frac{6 + 5}{600} = \frac{11}{600}$

$\Rightarrow f = \frac{600}{11} cm$

Topic Question Set

Q 1 :

An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is 25 D. Focal length of each of the convex lens is [2024]

Q 3 :

A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be: [2024]

Q 4 :

For the thin convex lens, the radii of curvature are 15 cm and 30 cm respectively. The focal length of the lens is 20 cm. The refractive index of the material is [2024]

Q 5 :

The position of the image formed by the combination of lenses is [2024]

Q 6 :

The following figure represents two biconvex lenses $L_{1}$ and $L_{2}$ having focal length 10 cm and 15 cm respectively. The distance between $L_{1}$ and $L_{2}$ is [2024]

Q 7 :

The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is __________ cm. [2024]

Q 8 :

In an experiment to measure the focal length ( $f$ ) of a convex lens, the magnitude of object distance ( $x$ ) and the image distance ( $y$ ) are measured with reference to the focal point of the lens. The $y - x$ plot is shown in figure.

The focal length of the lens is _____ cm. [2024]

Q 9 :

Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at _____ cm from the object. [2024]

Q 10 :

The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is ______ cm. [2024]

Q 11 :

In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to [ $R_{1}$ ] and [ $R_{2}$ ], i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is [2025]

Q 12 :

Given is a thin convex lens of glass (refractive index $μ$ ) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself. [2025]

Q 13 :

A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4D then the power of a part of the divided lens is [2025]

Q 16 :

What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D? ['D' stands for dioptre] [2025]

Q 19 :

A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is 2 cm $\times$ 2 cm and the area of the landscape photographed is 400 ${km}^{2}$ . The focal length of the lens in the drone camera is [2025]

Q 21 :

Two identical symmetric double convex lenses of focal length f are cut into two equal parts $L_{1}, L_{2}$ by AB plane and $L_{3}, L_{4}$ by XY plane as shown in figure respectively. The ratio of focal length of lenses $L_{1}$ and $L_{3}$ is [2025]

Q 22 :

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is: [2025]

Q 23 :

A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to [2025]

Q 24 :

A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is: [2025]

Q 25 :

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

Q 27 :

The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm. The refractive index of the lens material is [2025]

Q 28 :

A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when it is placed in water.

(Take refractive index of water as 1.28) [2025]

Q 29 :

Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is: [2025]

Q 30 :

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be [2025]

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Topic Question Set

Q 1 : An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is 25 D. Focal length of each of the convex lens is [2024]

Q 3 : A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be: [2024]

Q 4 : For the thin convex lens, the radii of curvature are 15 cm and 30 cm respectively. The focal length of the lens is 20 cm. The refractive index of the material is [2024]

Q 5 : The position of the image formed by the combination of lenses is [2024]

Q 6 : The following figure represents two biconvex lenses L1 and L2 having focal length 10 cm and 15 cm respectively. The distance between L1 and L2 is [2024]

Q 7 : The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is __________ cm. [2024]

Q 8 : In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The y-x plot is shown in figure. The focal length of the lens is _____ cm. [2024]

Q 9 : Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at _____ cm from the object. [2024]

Q 10 : The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is ______ cm. [2024]

Q 11 : In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to [R1] and [R2], i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is [2025]

Q 12 : Given is a thin convex lens of glass (refractive index μ) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself. [2025]

Q 13 : A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4D then the power of a part of the divided lens is [2025]

Q 16 : What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D? ['D' stands for dioptre] [2025]

Q 19 : A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is 2 cm × 2 cm and the area of the landscape photographed is 400 km2. The focal length of the lens in the drone camera is [2025]

Q 21 : Two identical symmetric double convex lenses of focal length f are cut into two equal parts L1,L2 by AB plane and L3,L4 by XY plane as shown in figure respectively. The ratio of focal length of lenses L1 and L3 is [2025]

Q 22 : Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is: [2025]

Q 23 : A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to [2025]

Q 24 : A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is: [2025]

Q 25 : A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

Q 26 : A bi-convex lens has radius of curvature of both the surfaces same as 1/6 cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides (R1≠R2), without any change in lens power then possible combination of R1 and R2 is : [2025]

Q 27 : The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm. The refractive index of the lens material is [2025]

Q 28 : A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28) [2025]

Q 29 : Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is: [2025]

Q 30 : A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be [2025]

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Q 1 :

An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is 25 D. Focal length of each of the convex lens is [2024]

Q 3 :

A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be: [2024]

Q 4 :

For the thin convex lens, the radii of curvature are 15 cm and 30 cm respectively. The focal length of the lens is 20 cm. The refractive index of the material is [2024]

Q 5 :

The position of the image formed by the combination of lenses is [2024]

Q 6 :

The following figure represents two biconvex lenses $L_{1}$ and $L_{2}$ having focal length 10 cm and 15 cm respectively. The distance between $L_{1}$ and $L_{2}$ is [2024]

Q 7 :

The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is __________ cm. [2024]

Q 8 :

In an experiment to measure the focal length ( $f$ ) of a convex lens, the magnitude of object distance ( $x$ ) and the image distance ( $y$ ) are measured with reference to the focal point of the lens. The $y - x$ plot is shown in figure.

The focal length of the lens is _____ cm. [2024]

Q 9 :

Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at _____ cm from the object. [2024]

Q 10 :

The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is ______ cm. [2024]

Q 11 :

In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to [ $R_{1}$ ] and [ $R_{2}$ ], i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is [2025]

Q 12 :

Given is a thin convex lens of glass (refractive index $μ$ ) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself. [2025]

Q 13 :

A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4D then the power of a part of the divided lens is [2025]

Q 16 :

What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D? ['D' stands for dioptre] [2025]

Q 19 :

A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is 2 cm $\times$ 2 cm and the area of the landscape photographed is 400 ${km}^{2}$ . The focal length of the lens in the drone camera is [2025]

Q 21 :

Two identical symmetric double convex lenses of focal length f are cut into two equal parts $L_{1}, L_{2}$ by AB plane and $L_{3}, L_{4}$ by XY plane as shown in figure respectively. The ratio of focal length of lenses $L_{1}$ and $L_{3}$ is [2025]

Q 22 :

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is: [2025]

Q 23 :

A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to [2025]

Q 24 :

A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is: [2025]

Q 25 :

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is (C is the center of curvature of the spherical surface and R is the radius of curvature) [2025]

Q 27 :

The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm. The refractive index of the lens material is [2025]

Q 28 :

A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when it is placed in water.

(Take refractive index of water as 1.28) [2025]

Q 29 :

Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is: [2025]

Q 30 :

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be [2025]