Question

Light of wavelength $6000\stackrel{\circ }{A}$ is incident along the normal on a glass plate of refractive index 1.5. What should be the smallest thickness of the plate which will make it appear darker when reflection is observed ?

• A. $6000\stackrel{\circ }{A}$
• B. $5000\stackrel{\circ }{A}$
• C. $3000\stackrel{\circ }{A}$
• D. $2000\stackrel{\circ }{A}$

Answer 1

The correct option is D $2000\stackrel{\circ }{A}$

For rays incident along the normal on the glass slab, two kinds of reflection can occur, shown in the figure.


If ray 1 and 2 interfere destructively, plate will appear darker.
Condition for destructive interference is
$\Delta \varphi =(2n-1)\pi$

Ray 1 suffers reflection from a denser medium, hence phase change $=\pi$
Ray 2 suffers reflection from a rarer medium, hence there is no phase change.

Ray 2 has an extra optical path length of $2t\mu$ because it passes through glass slab. Phase difference due to this $=\frac{2\pi }{\lambda }\times \left(2t\mu \right)$
Hence, net phase difference between 1 & 2
$\Delta \varphi =\frac{2\pi }{\lambda }\times \left(2t\mu \right)-\pi$

Now, $\Delta \varphi =(2n-1)\pi$
$\Rightarrow (2n-1)\pi =\frac{2\pi }{\lambda }\times \left(2t\mu \right)-\pi$
For smallest value of $t$, we should take $n=1$
$\Rightarrow \pi =\frac{2\pi }{\lambda }\times \left(2t\mu \right)-\pi$
$\Rightarrow t=\frac{{\lambda }_0}{2\mu }=\frac{6000\stackrel{\circ }{A}}{2\times \frac{3}{2}}=2000\stackrel{\circ }{A}$