Question

Young's double slit experiment is carried with two sheets of thickness 10.4 $\mu a$ each and refractive index $\mu =1.52and{\mu }_2=1.40$ covering the slits $s_{1}and{s}_2$ respectively. If while light of range 400 nm to 780 nm is used then which wavelength will from maximum exactly at pointO the center of the screen?

• A. 416 nm only
• B. 624 nm only
• C. 416 nm and 624 nm only
• D. none of these

Answer 1

The correct option is C 416 nm and 624 nm only

Given,

Thickness $t=10.4\mu m$

${\mu }_1=1.52{\mu }_2=1.40$

Range of light is $400nm-780nm$

Path difference is

$△x_1=({\mu }_1-1)t△x_2=({\mu }_2-1)t$

$△x=△x_1-△x_2=({\mu }_1-1)t-({\mu }_2-1)t=({\mu }_1-{\mu }_2)t$

Putting the value of refractive index and thickness in above equation

$△x=(1.52-1.4)\times 10.4\times {10}^{-6}m=1.248\times {10}^{-6}m=1248nm$

And path difference is $n\lambda$

Now

$n\lambda =1248nm\lambda =\frac{1248nm}{n}$

Range is 400 to 780nm

Therefore,

if $n=1$

$\lambda =\frac{1248nm}{1}=1248nm$

if $n=2$

$\lambda =\frac{1248nm}{2}=624nm$

if $n=3$

$\lambda =\frac{1248nm}{3}=416nm$

$416nm$ and $624nm$ is within range Hence, correct option is C