Wave Optics JEE Advanced PYQs | Physics Previous Year Questions & Answers

Q 1 :

In a Young's double slit experiment, the slit separation $d$ is 0.3 mm and the screen distance $D$ is 1 m. A parallel beam of light of wavelength 600 nm is incident on the slits at angle $α$ as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm. Which of the following statement(s) is/are correct? [2019]

[IMAGE 1154]

For $α = 0$ , there will be constructive interference at point $P$
Fringe spacing depends on $α$
For $α = \frac{0.36}{π}$ degree, there will be destructive interference at point $P$
For $α = \frac{0.36}{π}$ degree, there will be destructive interference at point $O$

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(3)

Path difference, $Δ x = d \sin α + d \sin θ = d α + \frac{y d}{D}$ $[when α and θ are small]$

[IMAGE 1155]

(1) For $α = 0$ , path difference

$Δ x = \frac{y d}{D} = \frac{0.3 \times 11}{1000} = 33 \times 10^{- 4} mm$

Now, $\frac{Δ x}{λ} = \frac{33 \times 10^{- 4}}{600 \times 10^{- 6}} = \frac{11}{2}$

$∴$ $Δ x = \frac{11}{2} λ = (2 n - 1) \frac{λ}{2}$

Hence, interference at $P$ is destructive.

(2) Fringe width, $β = \frac{λ D}{d}$ is independent of $α$

(3) For $α = \frac{0.36}{π}$ degree (at point P),

$Δ x = d [α + \frac{y}{D}]$ $= 0.3 \times 10^{- 3} [\frac{0.36}{180} + \frac{11 \times 10^{- 3}}{1}] m = 3900 nm$

Now, $\frac{Δ x}{λ} = \frac{3900}{600} = \frac{13}{2} λ$

Hence, destructive interference at $P$

(4) For $α = \frac{0.36}{π}$ degree (at point O)

$Δ x = d α = 0.3 \times 10^{- 3} \times \frac{0.36}{180} = 600 \times 10^{- 9} m = 600 nm$

Now, $\frac{Δ x}{λ} = 1$ $\Rightarrow Δ x = 1 λ$

Hence, constructive interference at $O$

Q 2 :

In the Young's double slit experiment using a monochromatic light of wavelength $λ$ , the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is [2013]

$(2 n + 1) \frac{λ}{2}$
$(2 n + 1) \frac{λ}{4}$
$(2 n + 1) \frac{λ}{8}$
$(2 n + 1) \frac{λ}{16}$

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(2)

Intensity, $I = I_{0} \cos^{2} \frac{ϕ}{2}$ where $I_{0}$ is the peak intensity.

Here, $I = \frac{I_{0}}{2},$ $∴$ $\frac{I_{0}}{2} = I_{0} \cos^{2} \frac{ϕ}{2},$ $∴$ $ϕ = \frac{π}{2} (2 n + 1)$

$∴$ $ϕ = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \dots$

And path difference,

$Δ x = (\frac{λ}{2 π}) ϕ$

$∴$ $Δ x = \frac{λ}{4}, \frac{3 λ}{4}, \dots, \frac{(2 n + 1) λ}{4}$

Q 3 :

Young's double slit experiment is carried out by using green, red and blue light, one colour at a time. The fringe widths recorded are $b_{G}$ , $b_{R}$ and $b_{B}$ respectively. Then, [2012]

$b_{G} > b_{B} > b_{R}$
$b_{B} > b_{G} > b_{R}$
$b_{R} > b_{B} > b_{G}$
$b_{R} > b_{G} > b_{B}$

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(4)

We know that fringe width, $β = \frac{λ D}{d}$

$∵$ $λ_{R} > λ_{G} > λ_{B}$

$∴$ $β_{R} > β_{G} > β_{B}$

Q 4 :

In Young's double slit experiment intensity at a point is $(\frac{1}{4})$ of the maximum intensity. Angular position of this point is [2005]

$\sin^{- 1} (\frac{λ}{d})$
$\sin^{- 1} (\frac{λ}{2 d})$
$\sin^{- 1} (\frac{λ}{3 d})$
$\sin^{- 1} (\frac{λ}{4 d})$

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(3)

[IMAGE 1156]

$I = I_{\max} \cos^{2}$

$= I_{\max} \cos^{2} (\frac{π d \sin θ}{λ})$

$4 = \frac{2 π}{λ} d \sin θ$

$\Rightarrow \frac{I_{\max}}{4} = I_{\max} \cos^{2} (\frac{π d \sin θ}{λ})$

or, $\cos (\frac{π d \sin θ}{λ}) = \frac{1}{2}$ $\Rightarrow \frac{π d \sin θ}{λ} = \frac{π}{3}$

$∴$ $θ = \sin^{- 1} (\frac{λ}{3 d})$

Q 5 :

Monochromatic light of wavelength 400 nm and 560 nm are incident simultaneously and normally on a double slits apparatus whose slit separation is 0.1 mm and screen distance is 1 m. Distance between areas of total darkness will be [2004]

4 mm
5.6 mm
14 mm
28 mm

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(4)

At the area of total darkness, minima will occur for both the wavelengths incident simultaneously and normally.

$∴$ $\frac{(2 n + 1)}{2} λ_{1} = \frac{(2 m + 1)}{2} λ_{2}$ $\Rightarrow (2 n + 1) λ_{1} = (2 m + 1) λ_{2}$

or, $\frac{(2 n + 1)}{(2 m + 1)} = \frac{560}{400} = \frac{7}{5}$ or $10 n = 14 m + 2$

By inspection, for $m = 2, n = 3$ and for $m = 7, n = 10$ , the distance between them will be the distance between such points.

i.e., $Δ s = \frac{D λ_{1}}{d} [\frac{(2 n_{2} + 1) - (2 n_{1} + 1)}{2}]$

Put $n_{2} = 10, n_{1} = 3$

$Δ s = \frac{1 \times (400 \times 10^{- 9})}{0.1 \times 10^{- 3}} [\frac{21 - 7}{2}] = 28 mm$

Q 6 :

In the ideal double-slit experiment, when a glass plate (refractive index 1.5) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $λ$ ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is [2002]

$2 λ$
$\frac{2 λ}{3}$
$\frac{λ}{3}$
$λ$

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(1)

Path difference $= (μ - 1) t = n λ;$

For minimum $t$ , $n = 1;$ $∴$ $t = \frac{n λ}{(μ - 1)} = \frac{λ}{(1.5 - 1)} = 2 λ$

Q 7 :

In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by [2001]

12
18
24
30

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(2)

Fringe width, $ω = \frac{12 λ_{1} D}{d} = \frac{k λ_{2} D}{d}$

$\Rightarrow k = \frac{12 \times 600}{400} = 18$

Hence, the number of fringes observed in the same segment of the screen $= 18$

Q 8 :

In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern [2000]

the intensities of both the maxima and the minima increase
the intensity of the maxima increases and the minima has zero intensity
the intensity of the maxima decreases and that of the minima increases
the intensity of the maxima decreases and the minima has zero intensity

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(1)

When slits are of equal width,

$I_{\max} \propto {(a + a)}^{2} = (4 a^{2})$

$I_{\min} \propto {(a - a)}^{2} = (0)$

When one slit's width is twice that of the other,

$\frac{I_{1}}{I_{2}} = \frac{W_{1}}{W_{2}} = \frac{a^{2}}{b^{2}}$ $\Rightarrow \frac{W}{2 W} = \frac{a^{2}}{b^{2}}$ $\Rightarrow b = \sqrt{2} a$

$∴$ $I_{\max} \propto {(a + \sqrt{2} a)}^{2} = (5.8 a^{2})$

$I_{\min} \propto {(\sqrt{2} a - a)}^{2} = (0.17 a^{2})$

Clearly, the intensities of both the maxima and minima increase.

Q 9 :

A Young's double slit interference arrangement with slits $S_{1}$ and $S_{2}$ is immersed in water (refractive index $= \frac{4}{3}$ ) as shown in the figure.

[IMAGE 1157]

The positions of maximum on the surface of water are given by $x^{2} = p^{2} m^{2} λ^{2} - d^{2},$ where $λ$ is the wavelength of light in air (refractive index $= 1$ ), $2 d$ is the separation between the slits and $m$ is an integer. The value of $p$ is [2015]

(3)

For maxima, Path difference $= m λ$

$∴$ $S_{2} A - S_{1} A = m λ$

[IMAGE 1158]

$∴$ $[(n - 1) \sqrt{d^{2} + x^{2}} + \sqrt{d^{2} + x^{2}}] - \sqrt{d^{2} + x^{2}} = m λ$

$∴$ $(n - 1) \sqrt{d^{2} + x^{2}} = m λ$

$∴$ $(\frac{4}{3} - 1) \sqrt{d^{2} + x^{2}} = m λ$

$∴$ $\sqrt{d^{2} + x^{2}} = 3 m λ$

$∴$ $d^{2} + x^{2} = 9 m^{2} λ^{2}$

$∴$ $x^{2} = 9 m^{2} λ^{2} - d^{2}$

Comparing this equation with the given equation $x^{2} = p^{2} m^{2} λ^{2} - d^{2},$

we get $p^{2} = 9$

$∴$ $p = 3$

Q 10 :

A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index $n_{2}$ . The other slit is at the interface of this medium with another medium 1 of refractive index $n_{1} (n_{1} \neq n_{2})$ . The line joining the slits is perpendicular to the interface and the distance between the slits is $d$ . The slit widths are much smaller than $d$ . A monochromatic parallel beam of light is incident on the slits from medium 1. A detector is placed in medium 2 at a large distance from the slits, and at an angle $θ$ from the line joining them, so that $θ$ equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.

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Which of the following statement(s) is(are) correct? [2022]

The phase difference between the two rays is independent of $d .$
The two rays interfere constructively at the detector.
The phase difference between the two rays depends on $n_{1}$ but is independent of $n_{2}$ .
The phase difference between the two rays vanishes only for certain values of $d$ and the angle of incidence of the beam, with $θ$ being the corresponding angle of refraction.

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Select one or more options

(1, 2)

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We have, $A B = d \tan θ$

and $B C = A B \sin α = d \tan θ \sin α$

Also, $A D = A B \sin θ = d \tan θ \sin θ$

So, optical path difference

$= n_{1} B C - n_{2} A D$

$= n_{1} (d \tan θ \sin α) - n_{2} (d \tan θ \sin θ)$

$= d \tan θ (n_{1} \sin α - n_{2} \sin θ)$ $= d \tan θ \times 0 = 0$

So, (1) and (2) are correct, and (3) and (4) are incorrect.

Q 11 :

While conducting the Young's double slit experiment, a student replaced the two slits with a large opaque plate in the $x - y$ plane containing two small holes that act as two coherent point sources $(S_{1}, S_{2})$ emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the $x - z$ plane (for $z > 0$ ) at a distance D = 3 m from the mid-point of $S_{1} S_{2}$ , as shown schematically in the figure. The distance between the sources is $d = 0.6003 mm$ . The origin O is at the intersection of screen and the line joining $S_{1} S_{2}$ . Which of the following is(are) true of the intensity pattern on the screen? [2016]

[IMAGE 1161]

Straight bright and dark bands parallel to the x-axis
The region very close to the point O will be dark
Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction
Semicircular bright and dark bands centered at point O

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Select one or more options

(2, 4)

Path difference at $O$ $= d = 0.6003 mm$

For $n \frac{λ}{2} = d,$

we get $n = 2001$

As $n$ is a whole number, the condition for minima is satisfied.

Therefore, $' O'$ will be dark, i.e., minima is formed at $' O'$ .

Also, as the screen is perpendicular to the plane containing the slits $S_{1}$ and $S_{2}$ , the fringes obtained will be semi-circular (only the top half of the screen is available)

Q 12 :

A light source, which emits two wavelengths $λ_{1} = 400 nm$ and $λ_{2} = 600 nm$ , is used in a Young's double slit experiment. If recorded fringe widths for $λ_{1}$ and $λ_{2}$ are $β_{1}$ and $β_{2}$ and the number of fringes for them within a distance $y$ on one side of the central maximum are $m_{1}$ and $m_{2}$ respectively, then [2014]

$β_{2} > β_{1}$
$m_{1} > m_{2}$
From the central maximum, $3^{rd}$ maximum of $λ_{2}$ overlaps with $5^{th}$ minimum of $λ_{1}$
The angular separation of fringes for $λ_{1}$ is greater than that for $λ_{2}$

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Select one or more options

(1, 2, 3)

We know that fringe width, $β = \frac{λ D}{d}$

$∵$ $λ_{2} > λ_{1}$

$∴$ $β_{2} > β_{1}$

Number of fringes in a given width, $m \propto \frac{1}{β}$

$∴$ $m_{1} > m_{2}$

$3 \times \frac{λ_{2} D}{d} = \frac{(2 \times 5 - 1) λ_{1} D}{2 d}$

$3 \times 600 = 4.5 \times 400$

Angular separation, $\frac{λ}{d} \propto λ$

So, it is greater for $λ_{2}$

Q 13 :

In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $λ$ . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). [2008]

If $d = λ$ , the screen will contain only one maximum
If $λ < d < 2 λ$ , at least one more maximum (besides the central maximum) will be observed on the screen
If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase
If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase

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Select one or more options

(1, 2)

[IMAGE 1162]

Condition to obtain maxima in Young's double slit experiment is

$d \sin θ = n λ$ where $n$ is an integer.

(1) When $d = λ$

$λ \sin θ = n λ$ $\Rightarrow \sin θ = n$

When $n = 0, θ = 0$

When $n = 1, θ = 90 °$

(This will be a point on the screen which will be at infinity and therefore not practical.)

Other values of $n$ are invalid as $- 1 \leq \sin θ \leq 1 .$

Hence, only one maxima when $d = λ .$

(2) When $λ < d < 2 λ$

$\Rightarrow λ < \frac{n λ}{\sin θ} < 2 λ$ $(∵ d = \frac{n λ}{\sin θ})$

$\Rightarrow 1 < \frac{n}{\sin θ} < 2$

$∴$ $Possible values of n are 0, + 1, - 1$

Hence, there is at least one more maxima besides the central maxima.

As we know,

$I_{\max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$ and $I_{\min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$

Initially, $I_{1} = 4 I$ and $I_{2} = I$

$∴$ $I_{\max} = 9 I$ and $I_{\min} = I$

When $I_{1} = I_{2} = I$ then $I_{\max} = 4 I$ and $I_{\min} = 0$

i.e., when the intensities become equal, $I_{\min}$ reduces to zero.

Q 14 :

Column-I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits $S_{1}$ and $S_{2}$ . In each of these cases $S_{1} P_{0} = S_{2} P_{0}, S_{1} P_{1} - S_{2} P_{1} = \frac{λ}{4}$ and $S_{1} P_{2} - S_{2} P_{2} = \frac{λ}{3},$ where $λ$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $μ$ and thickness $t$ is pasted on slit $S_{2}$ . The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point $P$ on the screen from the two slits is denoted by $δ (P)$ and the intensity by $I (P)$ . Match each situation given in Column-I with the statement(s) in Column-II valid for that situation. [2009]

[IMAGE 1163]

A - r, s, t; B - t; C - p, s; D - q
A - p, s; B - q; C - t; D - r, s, t
A - r, s, t; B - t; C - q; D - p, s
A - r, s, t; B - q; C - t; D - p, s

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(2)

For path difference $λ / 4$ , phase difference is $\frac{π}{2} .$

For path difference $λ / 3$ , phase difference is $\frac{2 π}{3} .$

Here, $S_{1} P_{0} - S_{2} P_{0} = 0$

$∴$ $δ (P_{0}) = 0$

The path difference for $P_{1}$ and $P_{2}$ will not be zero. The intensities at $P_{0}$ is maximum.

Intensity continuously decreases from $P_{0}$ towards $P_{2}$ .

$∴$ $I (P_{0}) > I (P_{1})$

(B) At $P_{1}$ , path difference is zero. Hence $P_{1}$ is the brightest central fringe and $δ P_{1} = 0 .$

$δ P_{0} = \frac{λ}{4}, δ P_{1} = 0, δ P_{2} = \frac{λ}{12}$

(C) Here, $δ (P_{0}) = - \frac{λ}{2}; δ (P_{1}) = - \frac{λ}{4}; δ (P_{2}) = - \frac{λ}{6}$

$I (P_{0}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- π)$

$= I_{1} + I_{2} - 2 \sqrt{I_{1}} \sqrt{I_{2}} = I_{0} + I_{0} - 2 I_{0} = 0$

$I (P_{1}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- \frac{π}{2})$

$= I_{1} + I_{2} = I_{0} + I_{0} = 2 I_{0}$

$I (P_{2}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- \frac{π}{3})$

$= I_{1} + I_{2} + \sqrt{I_{1}} \sqrt{I_{2}} = I_{0} + I_{0} + I_{0} = 3 I_{0}$

$∴$ $I (P_{2}) > I (P_{1}) > I (P_{0})$

(D) Here, $δ P_{0} = \frac{3 λ}{4}; δ P_{1} = - \frac{λ}{2}; δ P_{2} = - \frac{5 λ}{12}$

$I (P_{0}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- \frac{3 π}{2})$

$= I_{1} + I_{2} = I_{0} + I_{0} = 2 I_{0}$

$I (P_{1}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- π)$

$= I_{1} + I_{2} - 2 \sqrt{I_{1}} \sqrt{I_{2}} = I_{0} + I_{0} - 2 \sqrt{I_{0}} \sqrt{I_{0}} = 0$

$I (P_{2}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- \frac{5 π}{6})$

$= I_{1} + I_{2} - \sqrt{3} \sqrt{I_{1}} \sqrt{I_{2}} = (2 - \sqrt{3}) I_{0}$

Clearly, $I (P_{1}) = 0$

$I (P_{0}) > I (P_{1})$ and $I (P_{2}) > I (P_{1})$

Q 15 :

In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm. The distance between the slits at time $t$ is given by $d = (0.8 + 0.04 \sin ω t) mm,$ where $ω = 0.08 {rad s}^{- 1}$ . The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is $6000 Å$ . The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O. [2024]

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Q. The $8^{th}$ bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer ( $μ$ m), is ______.

(601.50)

Fringe width for the $n^{th}$ fringe $y = n (\frac{λ D}{d})$

For the $8^{th}$ fringe, $y = 8 \frac{λ D}{d}$

$y_{\max} = 8 \frac{λ D}{d_{\min}}$ and $y_{\min} = 8 \frac{λ D}{d_{\max}}$

$y_{\max} - y_{\min} = 8 λ D [\frac{1}{d_{\min}} - \frac{1}{d_{\max}}]$

Given, $D = 1 m, λ = 6000 \times 10^{- 10} m$

$d_{\max} = 0.84 mm$

$d_{\min} = 0.76 mm$

$∴$ $y_{\max} - y_{\min} = 8 \times 6000 \times 10^{- 10} \times 1 [\frac{1}{0.76 \times 10^{- 3}} - \frac{1}{0.84 \times 10^{- 3}}]$

$= 8 \times 6 \times 10^{- 4} [\frac{0.08}{0.76 \times 0.84}]$ $= 6.015 \times 10^{- 4} m = 601.5 μ m$

Q 16 :

In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm. The distance between the slits at time $t$ is given by $d = (0.8 + 0.04 \sin ω t) mm,$ where $ω = 0.08 {rad s}^{- 1}$ . The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is $6000 Å$ . The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O. [2024]

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Q. The maximum speed in $μ m/s$ at which the $8^{th}$ bright fringe will move is _______.

(24)

Speed, $v = \frac{d y}{d t} = - n \cdot \frac{λ D}{d^{2}} \cdot \frac{d (d)}{d t}$ $[∵ y = n \frac{λ D}{d}]$

$d = 0.8 + 0.04 \sin ω t$ $(given)$

$\frac{d (d)}{d t} = 0.04 ω \cos ω t$

For $V_{\max}$ , $\frac{d (d)}{d t} ⟶ \max$

For $\frac{d (d)}{d t} ⟶ \max,$

$\cos ω t = 1$ $\Rightarrow \sin ω t = 0$

$\Rightarrow {(\frac{d (d)}{d t})}_{\max} = 0.04 ω$

$\Rightarrow d = 0.8 mm$

$∴$ $V_{\max} = \frac{8 \times 6000 \times 10^{- 10} \times 1 \times 0.04 \times 0.08}{0.8 \times 0.8 \times 10^{- 6} \times 10^{- 3}}$ $= 24 μ m/s .$

Topic Question Set

Q 2 :

In the Young's double slit experiment using a monochromatic light of wavelength $λ$ , the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is [2013]

Q 3 :

Young's double slit experiment is carried out by using green, red and blue light, one colour at a time. The fringe widths recorded are $b_{G}$ , $b_{R}$ and $b_{B}$ respectively. Then, [2012]

Q 4 :

In Young's double slit experiment intensity at a point is $(\frac{1}{4})$ of the maximum intensity. Angular position of this point is [2005]

Q 5 :

Monochromatic light of wavelength 400 nm and 560 nm are incident simultaneously and normally on a double slits apparatus whose slit separation is 0.1 mm and screen distance is 1 m. Distance between areas of total darkness will be [2004]

Q 7 :

In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by [2001]

Q 8 :

In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern [2000]

Q 13 :

In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $λ$ . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). [2008]

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Topic Question Set

Q 2 : In the Young's double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is [2013]

Q 3 : Young's double slit experiment is carried out by using green, red and blue light, one colour at a time. The fringe widths recorded are bG, bR and bB respectively. Then, [2012]

Q 4 : In Young's double slit experiment intensity at a point is (14) of the maximum intensity. Angular position of this point is [2005]

Q 5 : Monochromatic light of wavelength 400 nm and 560 nm are incident simultaneously and normally on a double slits apparatus whose slit separation is 0.1 mm and screen distance is 1 m. Distance between areas of total darkness will be [2004]

Q 7 : In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by [2001]

Q 8 : In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern [2000]

Q 13 : In a Young's double slit experiment, the separation between the two slits is d and the wavelength of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). [2008]

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Q 2 :

In the Young's double slit experiment using a monochromatic light of wavelength $λ$ , the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is [2013]

Q 3 :

Young's double slit experiment is carried out by using green, red and blue light, one colour at a time. The fringe widths recorded are $b_{G}$ , $b_{R}$ and $b_{B}$ respectively. Then, [2012]

Q 4 :

In Young's double slit experiment intensity at a point is $(\frac{1}{4})$ of the maximum intensity. Angular position of this point is [2005]

Q 5 :

Monochromatic light of wavelength 400 nm and 560 nm are incident simultaneously and normally on a double slits apparatus whose slit separation is 0.1 mm and screen distance is 1 m. Distance between areas of total darkness will be [2004]

Q 7 :

In a Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen is given by [2001]

Q 8 :

In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern [2000]

Q 13 :

In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $λ$ . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). [2008]