Q: In Young's experiment, \lambda = 600\text{ nm}, D = 1\text{ m} and d = 1\text{ mm}. The fringe width is:

0.6\text{ mm}. \beta = \dfrac{\lambda D}{d} = \dfrac{600 \times 10^{-9} \times 1}{10^{-3}} = 6 \times 10^{-4}\text{ m} = 0.6\text{ mm}.

Q: In Young's experiment, \lambda = 500\text{ nm}, D = 2\text{ m} and d = 0.5\text{ mm}. The fringe width is:

2\text{ mm}. \beta = \dfrac{500 \times 10^{-9} \times 2}{0.5 \times 10^{-3}} = 2 \times 10^{-3}\text{ m} = 2\text{ mm}.

Question 1

In Young's double slit experiment, the fringe width is given by:

Accepted Answer

\beta = \dfrac{\lambda D}{d}. Here D is the slit-to-screen distance and d the slit separation; widening the slits narrows the fringes.

Question 2

The central fringe in Young's double slit experiment is:

Accepted Answer

Bright, since the path difference there is zero. The centre of the screen is equidistant from both slits, so the waves arrive exactly in step.

Question 3

In Young's experiment with monochromatic light, the fringe width is:

Accepted Answer

The same for all orders of fringes. Successive maxima are separated by \dfrac{\lambda D}{d} regardless of order, so the fringes are evenly spaced.

Question 4

If white light is used in Young's double slit experiment, the pattern shows:

Accepted Answer

A white central fringe with coloured fringes on either side. Zero path difference is achromatic, so all colours reinforce at the centre. Away from it, each wavelength has its own fringe width, producing colour.

Question 5

If the entire Young's double slit apparatus is immersed in water of refractive index \dfrac{4}{3}, the fringe width will:

Accepted Answer

Decrease to three-fourths of its original value. The wavelength in water is \dfrac{\lambda}{\mu}, so \beta' = \dfrac{\beta}{\mu} = \dfrac{3}{4}\beta.

Question 6

Increasing the separation between the two slits, with everything else unchanged, causes the fringe width to:

Accepted Answer

Decrease. \beta = \dfrac{\lambda D}{d} varies inversely with d, so wider slit separation crowds the fringes together.

Question 7

In Young's double slit experiment, replacing red light with violet light makes the fringes:

Accepted Answer

Narrower and more closely spaced. Violet has the shorter wavelength, and since \beta \propto \lambda the fringe width falls.

Question 8

The distance of the n^{	ext{th}} bright fringe from the centre in Young's experiment is:

Accepted Answer

\dfrac{n\lambda D}{d}. Bright fringes need a path difference of n\lambda, which places them at y_n = n\beta = \dfrac{n\lambda D}{d}.

Question 9

In Young's experiment, \lambda = 600	ext{ nm}, D = 1	ext{ m} and d = 1	ext{ mm}. The fringe width is:

Accepted Answer

0.6	ext{ mm}. \beta = \dfrac{\lambda D}{d} = \dfrac{600 	imes 10^{-9} 	imes 1}{10^{-3}} = 6 	imes 10^{-4}	ext{ m} = 0.6	ext{ mm}.

Question 10

In Young's experiment, \lambda = 500	ext{ nm}, D = 2	ext{ m} and d = 0.5	ext{ mm}. The fringe width is:

Accepted Answer

2	ext{ mm}. \beta = \dfrac{500 	imes 10^{-9} 	imes 2}{0.5 	imes 10^{-3}} = 2 	imes 10^{-3}	ext{ m} = 2	ext{ mm}.

Question 11

In Young's double slit experiment, if the slit separation is halved and the screen distance doubled, the fringe width becomes:

Accepted Answer

Four times as large. \beta \propto \dfrac{D}{d}, so doubling D and halving d together multiply the fringe width by 4.

Question 12

In a biprism experiment, 20 fringes are seen in a field of view with light of 6000\,	ext{Å}. With light of 4000\,	ext{Å} in the same field, the number of fringes is:

Accepted Answer

30. The field width is fixed: N_1\beta_1 = N_2\beta_2, and \beta \propto \lambda, so 20 	imes 6000 = N_2 	imes 4000, giving N_2 = 30.

Question 13

The ratio of the fringe widths produced by violet light (400	ext{ nm}) and red light (700	ext{ nm}) in the same apparatus is:

Accepted Answer

4 : 7. \beta \propto \lambda, so the ratio is simply 400 : 700 = 4 : 7.

Question 14

In Young's experiment with \lambda = 600	ext{ nm}, D = 1	ext{ m} and d = 0.3	ext{ mm}, the distance of the third bright fringe from the centre is:

Accepted Answer

6	ext{ mm}. y_3 = \dfrac{3\lambda D}{d} = \dfrac{3 	imes 600 	imes 10^{-9} 	imes 1}{3 	imes 10^{-4}} = 6 	imes 10^{-3}	ext{ m}.

Question 15

The fringe width in an interference pattern is 0.5	ext{ mm}. The distance between the second and the fifth bright fringes is:

Accepted Answer

1.5	ext{ mm}. There are three fringe widths between them: 3 	imes 0.5 = 1.5	ext{ mm}.

Question 16

In Young's experiment, \beta = 0.4	ext{ mm}, D = 1.2	ext{ m} and d = 1.5	ext{ mm}. The wavelength of light used is:

Accepted Answer

500	ext{ nm}. \lambda = \dfrac{\beta d}{D} = \dfrac{0.4 	imes 10^{-3} 	imes 1.5 	imes 10^{-3}}{1.2} = 5 	imes 10^{-7}	ext{ m}.

Question 17

The path difference at the second dark fringe in Young's experiment is:

Accepted Answer

\dfrac{3\lambda}{2}. Dark fringes need (2n-1)\dfrac{\lambda}{2}. For the second dark fringe n = 2, giving \dfrac{3\lambda}{2}.

Question 18

In Young's experiment with \lambda = 500	ext{ nm}, D = 1	ext{ m} and d = 1	ext{ mm}, the width occupied by 10 fringes is:

Accepted Answer

5	ext{ mm}. \beta = \dfrac{500 	imes 10^{-9}}{10^{-3}} = 0.5	ext{ mm}, so 10 fringes span 5	ext{ mm}.

Question 19

In Young's double slit experiment, doubling the distance between the slits and the screen makes the fringe width:

Accepted Answer

Twice as large. \beta \propto D, so doubling the screen distance doubles the fringe width.

Question 20

In Young's experiment, \lambda = 6000\,	ext{Å}, d = 0.2	ext{ mm} and D = 1	ext{ m}. The fringe width is:

Accepted Answer

3	ext{ mm}. \beta = \dfrac{6 	imes 10^{-7} 	imes 1}{2 	imes 10^{-4}} = 3 	imes 10^{-3}	ext{ m} = 3	ext{ mm}.

Young's Double Slit Experiment — MH-CET Physics MCQs with Solutions

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