Question

In a standard $YDSE$ apparatus a thin film $(\mu =1.5,t=2.1\mu m)$ is placed in front of upper slit: How far above and below the centre point of the screen are two nearest maxima located ? Take $D=1m,d=1mm\lambda =4500\stackrel{\circ }{A}$. (Symbols have usual meaning).

• A. $1.5mm$
• B. $0.6mm$
• C. $0.15mm$
• D. $0.3mm$

Answer 1

The correct option is A $1.5mm$

Given,

$\mu =1.5t=2.1\mu mD=1md=1mm\lambda =4500A^{\circ }$

Fringe width is

$\beta =\frac{\lambda D}{d}$

Substitute all value in above equation

$\beta =\frac{4.5\times {10}^{-7}\times 1}{1\times {10}^{-3}}=0.45mm$

Path difference is

$pd=(\mu -1)t=(1.5-1)2.1\times {10}^{-6}=0.5\times 2.1\times {10}^{-6}=1.05\times {10}^{-6}\frac{dx}{D}=1.05\times {10}^{-6}x=\frac{1.05\times {10}^{-6}}{{10}^{-3}}=1.05mm$

Hence above the central point of the screen nearest maxima will be at

$0.45+1.05=1.5mm$

Correct option is A.