Question

In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of a wavelength 700 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 would be
1. 28
2. 24
3. 30
4. 18

Answer 1

Hint: In this question, we are given that in Young’s double slit experiment 16 fringes are observed in a certain segment of the screen where the wavelength is 700 nm, and we have to number of fringes when the wavelength is changed to 400 nm. Here, directly equate the young’s modulus for inference fringes of both the conditions and solve.

Formula used:
Young’s modulus for inference fringes –
$Y = \dfrac{{nD\lambda }}{d}$
Here,
$n = $ number of fringes
$\lambda = $ wavelength of the light
$d = $ distance between the sources (two slits)
$D = $ distance between the sources and the screen

Complete answer:
Given that,
${n_1} = 16$, ${\lambda _1} = 700nm$
${n_2} = ?$, ${\lambda _2} = 400nm$

As we know that, Young’s modulus for inference fringes is:
$Y = \dfrac{{nD\lambda }}{d}$

Therefore, for the young’s double slit experiment
${Y_1} = {Y_2}$
$\dfrac{{{n_1}D{\lambda _1}}}{d} = \dfrac{{{n_2}D{\lambda _2}}}{d}$
${n_1}{\lambda _1} = {n_2}{\lambda _2}$

It implies that,
${n_2} = \dfrac{{{n_1}{\lambda _1}}}{{{\lambda _2}}}$

Putting all the given values,
\[{n_2} = \dfrac{{16 \times 700}}{{400}}\]
\[{n_2} = 28\]

Hence, Option (1) is the correct answer i.e., 28.

Note: The key concept involved in solving this problem is the good knowledge of Young’s double slit experiment. Students must remember that Young's double-slit experiment employs two coherent light sources separated by a small distance, often a few orders of magnitude greater than the wavelength of light. Young's double-slit experiment aided in establishing the wave theory of light.