Question

In a $YDSE$ experiment, the distance between the slits & the screen is $100\text{cm}$. For a certain distance between the slits, an interference pattern is observed on the screen with the fringe width $0.25\text{mm}$. When the distance between the slits is increased by $\Delta d=1.2\text{mm}$, the fringe width decreased to $n=2/3$ of the original value. In the final position, a thin glass plate of refractive index $1.5$ is kept in front of one of the slits & the shift of central maximum is observed to be $20$ fringe width. Which of the options are correct ?

• A. The wavelength of wave is $600\text{nm}$
• B. The wavelength of wave is $800\text{nm}$
• C. The width of plate is $20\mu \text{m}$
• D. The width of plate is $24\mu \text{m}$

Answer 1

Answer: (A), (D)

The width of plate is $24\mu \text{m}$

Clearly , Fringe width ${\beta }_{\text{initial}}=\frac{\lambda D}{d_i}$

Substituting the data given in the question,

$0.25\times {10}^{-3}=\frac{\lambda }{d}\times 1$

$\Rightarrow \frac{\lambda }{d}=2.5\times {10}^{-4}........\left(1\right)$

Afterwards :

$\frac{2}{3}{\beta }_i=\frac{\lambda D}{(d+\Delta d)}$

$\Rightarrow \frac{\lambda }{d+\Delta d}=\frac{2}{3}\times 2.5\times {10}^{-4}$ ...(2)

Dividing (1) and (2)

$\frac{d+\Delta d}{d}=\frac{3}{2}\Rightarrow d=2\left(\Delta d\right)=2.4\text{mm}$

$\therefore \lambda =2.4\times 2.5\times {10}^{-7}=600\text{nm}$


For point $P$ :

Fringe shift $s=\frac{D}{d}(\mu -1)t$

$\Rightarrow 20\beta =\frac{D}{d}(\mu -1)t$

$\Rightarrow 20\lambda =(\mu -1)t$

$\Rightarrow t=\frac{20\lambda }{\mu -1}=\frac{20\times 600\times {10}^{-9}}{(1.5-1)}=24\mu \text{m}$