Question

If Young’s double-slit experiment is performed in water instead of air, then
(A) No fringes would be seen
(B) Fringe width would decrease
(C) Fringe width would increase
(D) Fringe width would remain unchanged

Answer 1

Hint: In Young’s double-slit experiment, the single beam is broken into the two sources. In Young's double-slit experiment, the fringe width(β) is directly proportional to the distance between slit and screen (D) and wavelength of the light source and inversely proportional to the width of the slit which is (d).

Formula used:
In Young's double-slit experiment the fringe width is given below.
$\beta = \dfrac{{D\lambda }}{d}$
Where ,d is the width of the slit, D is the distance between slit and screen and λ is the wavelength.

 Complete step by step answer:
The relation between the speed of light and wavelength is given below.
$c = \lambda \upsilon $ …………... (1)
Where,
c=velocity of light,
$\lambda $=wavelength of the light source,
 ϑ=frequency
Now we know that changing the medium frequency remains constant. Therefore, using equation (1) we can conclude that the speed of light is directly proportional to the wavelength of light. And we also know that the speed of light is maximum in vacuum or air. This means the speed of light decreases in water as a result wavelength also decreases in water as frequency remains unchanged.
Now let us use the formula for fringe width which is given below.
$\beta = \dfrac{{D\lambda }}{d}$ ……………... (2)
From the equation, we can see that fringe width is directly proportional to the wavelength. And from relation (1) we have already concluded that wavelength will decrease in water.
Therefore, changing the medium speed will decrease as a result wavelength will also decrease, and if wavelength decreases fringe width also decreases.
Hence, option (B) is the correct option.

Additional information:
Young’s double-slit experiment is proof of the wave nature of light.
The interference pattern is obtained by the superposition of light coming from the two slits.
Constructive interference: \[{\text{dsin}}\theta {\text{ = n}}\lambda \left( {{\text{for n = 0, 1, - 1, 2, - 2}} \ldots {\text{.}}} \right)\]
Destructive interference: \[{\text{dsin}}\theta {\text{ = (n + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{)}}\lambda\left( {{\text{for n = 0, 1, - 1, 2, - 2}} \ldots{\text{.}}} \right)\]
Where, d=distance between the slits
θ is the angle to the incident direction
n=order of interference

Note:
In Young's double-slit experiment, a coherent light source illuminates a plate which is having two parallel slits, and the light passing through the slits is observed on a screen behind the plate.
Due to the wave nature of light, the light waves passing through the two slits produce bright and dark bands on the screen.