Question

Light of wavelength $7500A^o$ is incidents on a thin glass plate $(\mu =1.5)$ so that the angle of refraction obtained is ${30}^0$. If the plate appears the dark then the minimum thickness of the plate will be

• A. $4000\sqrt{3}{A}^o$
• B. $\frac{8000}{\sqrt{3}}{A}^o$
• C. $\frac{5000}{\sqrt{3}}{A}^o$
• D. $1000\sqrt{3}{A}^o$

Answer 1

Answer: (C)

$\frac{5000}{\sqrt{3}}{A}^o$

Path difference between ray 1 and ray 2 is $2\mu dsin\left(r\right)$

For plate to appear dark , the rays must interfere destructively. Hence the path difference should be an even multiple of $\lambda$.

This is because there is an extra phase change of $\pi$ due to reflection of ray 1 from the above layer(reflection from a higher refractive index medium, which is not the case with ray 2).

Therefore, $2\mu d\sin r=n\lambda$

$d=\frac{n\lambda }{2\mu sin\left(r\right)}$

For minimum thickness, order(n)=1

Hence, $d=\frac{5000}{\sqrt{3}}{A}^{\circ }$