Question 54: Wave Optics

Question

In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is $2.4 \mu \mathrm{~m}$. If the experiment is carried out in another medium having refractive index 1.2 , the fringe width will be $\_\_\_\_$ $\mu \mathrm{m}$.

Accepted Answer

In YoungÃ¢ÂÂs double slit experiment, fringe width is given by
$$ \beta = \frac{\lambda D}{d} $$
where
$\lambda$ = wavelength in the medium,
$D$ = distance between slit and screen,
$d$ = distance between the slits.
When the experiment is performed in a medium of refractive index $\mu$, the wavelength becomes
$$ \lambda' = \frac{\lambda}{\mu} $$
So the new fringe width is
$$ \beta' = \frac{\lambda' D}{d} = \frac{1}{\mu}\frac{\lambda D}{d} $$
Thus,
$$ \beta' = \frac{\beta}{\mu} $$
Given:
$$ \beta = 2.4\ \mu m,\qquad \mu = 1.2 $$
Therefore,
$$ \beta' = \frac{2.4}{1.2} = 2\ \mu m $$
So, the correct answer is:
$$ \boxed{2\ \mu m} $$
Option B

Answer

1.2

Answer

2

Answer

2.4

Answer

2.88

Wave Optics

In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is 2.4 \mu \mathrm{~m}. If the experiment is carried out in another medium having refractive index 1.2 , the fringe width will be \_\_\_\_ \mu \mathrm{m}.

Choose the correct answer: