Question

In a $YDSE$, both slits produce equal intensities on the screen. A $100\%$ transparent thin film is placed in front of one of the slits. Now, the intensity on the centre becomes $75\%$ of the previous intensity. The wavelength of the light is $6000A^o$ and refractive index of glass is $1.5$ The minimum thickness of the glass slab is:

• A. $0.2\mu m$
• B. $0.3\mu m$
• C. $1.4\mu m$
• D. $1.6\mu m$

Answer 1

The correct option is D $1.6\mu m$

Given,

Wavelength of light $\lambda =6000A^o$

Refractive index of glass, $\mu =1.5$

Let, Intensity be $I$, Phase difference $\delta$ , Path difference $\Delta x$, thickness of film $t$, refractive index of film $\mu$

Maximum intensity without film $I_{max}=I_o+I_o+2\sqrt{I_o{I}_o}\cos 0^o=4I_o$

For, 0.75% of Maximum, $=0.75\times 4I_o=3I_o$

$3I_o=I_o+I_o+2\sqrt{I_o{I}_o}\cos \delta$

$\Rightarrow \cos \delta =\frac{1}{2}$

$\Rightarrow \delta =(2n+1)\frac{\pi }{3}$

From path difference and phase difference relation.

$\delta =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }(\mu -1)t$

$\Rightarrow (2n+1)\frac{\pi }{3}=\frac{2\pi }{\lambda }(\mu -1)t$

$\Rightarrow t=\frac{\lambda (2n+1)}{6(\mu -1)}=\frac{6000\times {10}^{-10}(2n+1)}{6(1.5-1)}$

$t=2\times {10}^{-7}(2n+1)$

$t=0.2\mu m,0.6\mu m,1\mu m,1.4\mu m,1.8\mu m$

So, $1.6\mu m$cannot be the film thickness.