Q 11 :

A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is          [2025]

  • 30

     

  • 45

     

  • 607

     

  • 15

     

(4)

Equivalent focal length,

1f=1f1+1f2

1f=130+1–20=2–360=–160 ⇒ f=–60 cm

Lens formula

1v–1u=1f

1v–1–20=1–60 ⇒v=–15 cm



Q 12 :

Light from a point source in air falls on a spherical glass surface (refractive index, μ = 1.5 and radius of curvature = 50 cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ________ m.           [2025]



(4)

μ2v–μ1u=μ2–μ1R

1.5200–1–x=1.5–150

1x=1100–3400

⇒x=400 cm=4 m



Q 13 :

When a beam of white light is allowed to pass through a convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called                          [2023]

  • scattering

     

  • chromatic aberration

     

  • spherical aberration

     

  • polarisation

     

(2)

The phenomena is known as chromatic aberration.

 



Q 14 :

A person has been using spectacles of power –1.0 D for distant vision and a separate reading glass of power 2.0 D. What is the least distance of distinct vision for this person      [2023]

  • 10 cm

     

  • 40 cm

     

  • 30 cm

     

  • 50 cm

     

(4)

1v-1u=1f, P=2D

⇒1f=2100cm-1

      1v-(-125)=2100

⇒1v=150-125 ⇒ v=-50 cm



Q 15 :

As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is 30 cm and refractive index of the material for both the lenses is 1.75. Both the lenses are placed at distance of 40 cm from each other. Due to the combination, the image of the object is formed at distance x = ____ cm, from concave lens.                     [2023]



(120)

1f1=(1.75-1)(-130) 

⇒ f1=-40 cm

1f2=(1.75-1)(130) ⇒ f2=40 cm

Image from L1 will be virtual and on the left of L1 at focal length 40 cm. So the object for L2 will be 80 cm from L2 which is 2f. Final image is formed at 80 cm from L2 on the right.

So x=120



Q 16 :

A convex lens of refractive index 1.5 and focal length 18 cm in air is immersed in water. The change in focal length of the lens will be ________ cm.

(Given refractive index of water =43)                   [2023]



(54)

1fH2O=(μgμH2O-1)(2R)=18(2R)=1(4fair)

⇒  fH2O=4fair=72 cm

So change in focal length=72-18=54 cm



Q 17 :

An object is placed on the principal axis of a convex lens of focal length 10 cm as shown. A plane mirror is placed on the other side of lens at a distance of 20 cm. The image produced by the plane mirror is 5 cm inside the mirror. The distance of the object from the lens is ________cm.                   [2023]



(30)

f=10 cm

1v-1u=1f

115-1-u=110

⇒  1u=110-115

On solving we get value of u as 30 cm.



Q 18 :

Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principal axis in rarer medium at a distance of 15 cm from the pole of the surface. The distance of image from the pole of the surface is ________cm.                     [2023]



(30)

μ2v-μ1u=μ2-μ1R

1.5v-1-15=1.5-130=160

1.5v+115=160

1.5v=160-115=-120

1.5v=-120 ⇒ v=-30 cm



Q 19 :

A point object, 'O' is placed in front of two thin symmetrical coaxial convex lenses L1 and L2 with focal length 24 cm and 9 cm respectively. The distance between two lenses is 10 cm and the object is placed 6 cm away from lens L1 as shown in the figure. The distance between the object and the image formed by the system of two lenses is __________ cm.              [2023]



(34)

From 1st lens  1v+16=124

1v=124-16=-18 ⇒ v=-8 cm

From 2nd lens  1v+118=19

1v=19-118=118 ⇒ v=18

So distance between object and its image =6+10+18=34 cm



Q 20 :

The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm. The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be x:1. The value of x is ____ .                 [2023]



(4)

P=(1.8-1)(120+120)  by lens maker's formula

P'=(1.81.5-1)(120+120)

Dividing  PP'=0.81.2-1=4



Q 21 :

Two convex lenses of focal length 20 cm each are placed coaxially with a separation of 60 cm between them. The image of the distant object formed by the combination is at ____ cm from the first lens.                      [2023]



(100)

f1=20 cm, f2=20 cm

1st refraction in L1(I1)

1v-1u=1f ⇒ 1v-1∞=1f ∴ v=f

2nd refraction in L2

I1→object, I2→image also,  u=-40 cm and f=20 cm

1v-1u=1f ⇒ 1v-1(-40)=120     ∴ v=40 cm

Correct Answer is 100.



Q 22 :

A bi convex lens of focal length 10 cm is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is ________ D.   [2023]



(5)

Let power of each part is P1, then

P1+P1=P=1f

2P1=10.1=10

P1=5 D



Q 23 :

A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is __________ mm.   [2026]



(280)

402=f14

⇒  f=280 mm



Q 24 :

A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification m1 when an object is placed 10 cm before the convex lens. Keeping the positions of the convex lens and object undisturbed, a gap of 1 cm is introduced between the lenses by moving the concave lens away, which leads to a change in magnification of the total lens system to m2. The value of |m1m2| is _________.           [2026]

  • 527

     

  • 32

     

  • 2527

     

  • 56

     

(4)

(Dropped)



Q 25 :

A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. The refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of curvature of the spherical surface. The value of x is ______ cm.         [2026]



(100)

μ2v-μ1u=μ2-μ1R⇒1.5v-1∞=1.5-150

V=150 cm

x→measure from center

x=V-R

=150-50=100 cm



Q 26 :

A biconvex lens is formed by using two thin plano-convex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be:   [2026]

  • −2.5

     

  • −2

     

  • +2

     

  • +2.5

     

(2)

1v-1u=1fnet=1f1+1f2

1v+130=(1.5-1)(115-1∞)+(1.2-1)(1∞+112)

1v+130=130+160

v=60

m=vu=60-30=-2



Q 27 :

Five persons P1,P2,P3,P4 and P5 recorded object distance (u) and image distance (v) using same convex lens having power +5D as (25,96), (30,62), (35,37), (45,35) and (50,32) respectively. Identify correct statement                   [2026]

  • Readings recorded by P3 and P2 persons are incorrect

     

  • Readings recorded by P3 person are incorrect

     

  • Readings recorded by P4 and P5 persons are incorrect

     

  • Readings recorded by all persons are correct

     

(2)

P=+5D

1f=5 ⇒ f=20 cm

⇒ If object is between f and 2f, image will be beyond 2f and magnified.

⇒ If object is beyond 2f, image will be between f and 2f and diminished.

Hence reading of P3 are incorrect.



Q 28 :

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ______ cm.  [2026]



(16)

m=ff+u

m1=-m2

ff-8=-ff-24

f-8=24-f

2f=32

f=16 cm



Q 29 :

The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index 1.7) are same. If the curvature of the plano-concave lens exactly matches with the curvature of the back surface of the biconvex lens, then the ratio of radii of curvature of front and back surface of the biconvex lens is ______.   [2026]

  • 5 : 12

     

  • 5 : 2

     

  • 12 : 5

     

  • 2 : 5

     

(2)

|PA|=|PB|

0.5(1R1+1R2)=0.7R2

5R1=2R2

R1R2=52



Q 30 :

A convex lens of refractive index 1.5 and focal length f=18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α×f. The value of α is ____.

(Refractive index of water = 43)       [2026]



(3)

1fair=(1.5-11)(1R1+1R2)

1fwater=(1.5-4343)(1R1+1R2)

⇒fwaterfair=0.50.5/4=4

⇒fwater-fair=3f



Q 31 :

A virtual erect image by a diverging lens is represented by (u, v, f are object distance, image distance and focal length respectively)

  •  

  •  

  •  

  •  

(1)



Q 32 :

The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is 40 cm. The area of the image is 9 times that of the square. The focal length of the lens is:

  • 36 cm

     

  • 27 cm

     

  • 60 cm

     

  • 30 cm

     

(4)

If side of object square=l

and side of image square=l'

From question, l'2l2=9 or l'l=3

i.e. magnification m=3,

u=-40 cm

v=3×40=120 cm

f=?

From formula, 1v-1u=1f⇒1120-1-40=1f

Or, 1f=1120+140=1+3120

∴ f=30 cm



Q 33 :

Refer the figure given below. μ1 and μ2 are refractive indices of air and lens material. The height of image will be ______ cm.           [2026]

  • 1

     

  • 0.5

     

  • 1.2

     

  • 0.25

     

(1)

n2v-n1u=n2-n1R⇒1.54v-1-40=1.54-1-20

1.54v=-0.5420-140=-0.54×2-140

v=-1.54×402.08=-61.62.08=-29.61 cm

m=hIh0=v/n2u/n1=n1vn2u=1×29.611.54×40

hI=2×29.611.54×40=0.96 cm≅1 cm



Q 34 :

A convex lens is made from glass material having refractive index of 1.4 with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is ______.          [2026]

  • 0.5

     

  • 2.5

     

  • 0.8

     

  • 1.25

     

(4)

1f=(n-1)(2R)

1f=(0.4)(2)·1R

fR=10.8=108=54=1.25



Q 35 :

A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses?         [2026]

  • behaves as concave lens if |fconvex|>|fconcave|

     

  • behaves as concave lens if |fconvex|<|fconcave|

     

  • behaves as convex lens if |fconvex|>|fconcave|

     

  • Focal length of the lens system will change if the position of two lenses are interchanged

     

(1)

1feq=1fconvex+1fconcave

fconcave is -ve

fconvex is +ve

If |fconcave|<|fconvex|

then, feq is negative.



Q 36 :

An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm, the final image is ______.               [2026]

  • virtual, formed at 7.5 cm right of lens Q, with a size bigger than that of AB

     

  • real, formed at 7.5 cm right of lens Q, with a size same than that of AB

     

  • formed at infinity

     

  • real, formed at 7 cm right of lens Q, with a size smaller than that of AB

     

(2)

For lens P: u=-15 cm

v1=ufu+f=-15×10-15+10=-150-5=+30

m1=30-15=-2

For lens Q: u=+15 cm

v2=ufu+f=15×1515+15=15×1530=7.5 cm

m2=7.515=12

m=m1m2=-2×12=-1



Q 37 :

A spherical interface lens of radius R separates two media of refractive indices 1 and 1.4 respectively as shown in the figure below. A point source is placed at a distance of 4R in front of spherical interface. The magnitude of the magnification of point source image is ______.          [2026]

  • 1.66

     

  • 2.33

     

  • 2.66

     

  • 1.33

     

(1)

μ2v-μ1u=μ2-μ1R

1.4v-1-4R=1.4-1R⇒v=283R

m=μ1μ2vu=11.4×28R3×4R=74.2=1.66



Q 38 :

A thin biconvex lens is prepared from the glass (μ = 1.5), both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of ______ cm.         [2026]

  • 10

     

  • 12.5

     

  • 13

     

  • 13.5

     

(1)

Combination of lens and the mirror

fm=R2=-10 cm;  fL=R2(μ-1)=20 cm

1f=1fm-2fL

1f=-110-220=-15⇒f=-5 cm

To form image at the position of object, it should be placed at center of curvature of mirror and the lens combination.

∴ u=2f=2×5=10 cm



Q 39 :

If sunlight is focused on a paper using convex lens, it starts burning the paper in shortest time when the lens is kept at 30 cm above the paper. If the radius of curvature of the lens is 60 cm then the refractive index of the lens material is α10. The value of α is ______.               [2026]



(20)

 To burn in minimum time distance of paper must be at focus.

So f=3 cm

Lens maker formula,

1f=(μ-1)(1R1-1R2)

130=(μ-1)(2R)

130=(μ-1)(260)

μ=2⇒α10=2⇒α=20