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With a monochromatic light, the fringe-width obtained in a young’s double-slit experiment is $0.133\;cm$ . The whole set-up is immersed in water of refractive index $1.33\;$ , then the new fringe- width is
(A) $0.133\;cm$
(B) $0.1\;cm$
(C) $1.33 \times 1.33cm$
(D) $\dfrac{{1.33}}{2}cm$

Last updated date: 13th Jun 2024
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Hint: To solve this type of question firstly we have to find all the variables which are useful for finding the new fringe-width. So we will find the wavelength of light in water. And all other quantities are already given. Note that whenever the light source is placed in a different medium of the relative refractive index of $n$, there is a change in wavelength.
Formula used:
The formula for fringe width in YDSE is given by:
$\beta = \dfrac{{\lambda D}}{d}$

Complete step-by-step solution:
In Young's double-slit experiment a coherent light source illuminates a plate stabbed by two parallel slits, and the light passing through the slits is detected on a screen behind the plate. The wave nature of light causes the light waves to pass on through the two slits to interfere, creating bright and dark bands on the screen. The width of each fringe formed has a definite value across the entire band and it is known by the below formula.
The formula for fringe width in a double’s slit experiment is known by:
$\beta = \dfrac{{\lambda D}}{d}$
$\lambda $ = wavelength of light
$D$ is the distance between the light source and the screen
$d$ is the distance between the slits.
Rearranging the equation, we can obtain
$\lambda = \dfrac{{\beta d}}{D}$ .......... $\left( 1 \right)$
When the whole experiment is conducted in water, there is a shift in the wavelength because whenever the light source is placed in a different medium of the refractive index of $n$ , there is a change in wavelength, given by:
$\lambda ' = \dfrac{\lambda }{n}$
Substitute the value of $\lambda $ in the equation $\left( 1 \right)$ , we have:
$\beta ' = \dfrac{{\lambda 'D}}{d}$
$\beta ' = \dfrac{{\beta d}}{{nD}} \times \dfrac{D}{d}$
On further solving the above equation we get,
$\beta ' = \dfrac{\beta }{n}$
Put the values that are already given in the question we get,
$\beta ' = \dfrac{\beta }{n}$
$ \Rightarrow \beta ' = \dfrac{{0.133}}{{1.33}} = 0.1cm$
Thus, the new fringe-width $\beta ' = 0.1cm$ .

Hence, the correct option is (B) $0.1\;cm$ .

Note: The wavelength of light in the denser medium will always be lesser than that of air since the refractive index of any denser medium like water, glass, etc. is greater than one. So, if you are doing the actual calculation with numbers, you should make sure that the wavelength in the denser medium should always be lesser, to ensure that you are moving in the right direction.