Question

In double slit experiment, a light of wavelength $\lambda =600nm$ is used. When a film of material $3.6\times {10}^{-3}$ cm thick was placed over one of the slits, the fringe pattern was displaced by a distance equal to 30 times that between two adjacent fringes. What is the refractive index of the material?

• A. 1.5
• B. 7.2
• C. 8.1
• D. 5.5

Answer 1

The correct option is A 1.5

$n=\frac{(\mu -1)fD}{d}$ (1)

$n=$ Where, no. of fringes

$\mu =$ refrachive index of material thickness of material

$D=$ distance $b/\omega$ screen and slit

$d=$ distance $b/\omega$ slit

$n=30\frac{\lambda D}{d}\to \text{(ii)}$

from

(1) and (11)

$\Rightarrow 30\frac{\lambda D}{d}=\frac{(\mu -1)tD}{d}$

$\Rightarrow 30\times 600\times {10}^{-9}=(\mu -1)3.6\times {10}^{-3}\times {10}^{-2}$

$\Rightarrow (\mu -1)=\frac{18\times {10}^{3-9}}{3.6\times {10}^{-5}}$

$\Rightarrow (\mu -1)=\frac{1}{2}\times {10}^{-6+5+1}$

$\Rightarrow (\mu -1)=0.5$

$\therefore \mu =1,5$

(A) 1.5 is correct