Question

A parallel beam of sodium light of wavelength $6000\mathring{A}$ is incident on a thin glass plate of $\mu =1.5$, such that the angle of refraction in the plate is ${60}^{\circ }$. The smallest thickness of the plate which will make it appear dark by reflected light is

• A. $40\mathring{A}$
• B. $4\mathring{A}$
• C. $400\mathring{A}$
• D. $4000\mathring{A}$

Answer 1

The correct option is D $4000\mathring{A}$

For reflected light, condition of destructive interference,

$2\mu t\cos r-\frac{\lambda }{2}=(2n+1)\frac{\lambda }{2}$

$\Rightarrow 2\mu t\cos r=(2n+1)\frac{\lambda }{2}+\frac{\lambda }{2}$

For minimum thickness of the plate, $n=0$,

So,

$2\mu t\cos r=\lambda$

$\Rightarrow 2\times 1.5\times t\times \cos {60}^{\circ }=6000\mathring{A}$

$\Rightarrow t=\frac{6000}{1.5}=4000\mathring{A}$

Hence, option $\left(D\right)$ is the correct answer.