Question

White light is incident normally on a glass plate $\left(\text{in air}\right)$ of thickness $500\text{nm}$ and refractive index of $1.5$. The wavelength $\left(\text{in nm}\right)$ in the visible region $\left(\text{400 nm-700 nm}\right)$ that is strongly reflected by the plate is

• A. $450$
• B. $600$
• C. $400$
• D. $500$

Answer 1

The correct option is B $600$


Ray$-1$ is reflected when it travels from a rarer medium to a denser medium, so it is subjected to a phase difference $\Delta \varphi =\pi$ due to reflection.

Since, ray$-2$ is reflected when it travels from a denser medium to a rarer medium. So, phase difference due to reflection will be zero.

So path difference will be,

$\Delta x=\frac{\lambda }{2\pi }\times \Delta \varphi =\frac{\lambda }{2}$

But, the optical path difference of ray $2$ is given by

$\Delta x=2\mu t$

For, maximum intensity,
$\Delta x=2\mu t=(2n-1)\frac{\lambda }{2}$

$\Rightarrow \lambda =\frac{4\mu t}{(2n-1)}$

$\Rightarrow \lambda =\frac{4\times 1.5\times 500\text{nm}}{2n-1}$

$\Rightarrow \lambda =\left(\frac{3000}{2n-1}\right)\text{nm}$

Substituting $n=1,2,3,.....etc$, we get
$\lambda =3000\text{nm},1000\text{nm},600\text{nm}...etc.$

$\therefore$ Only $600\text{nm}$ lies in the visible region.

Hence, option $\left(\text{B}\right)$ is correct