Question 178: Wave Optics

Question

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56 , the central fringe shifts to the position of $7^{	ext {th }}$ bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm , the thickness of mica sheet is $\alpha 	imes 10^{-9} \mathrm{~m}$. The value of $\alpha$ is
$\_\_\_\_$ .

Accepted Answer

In YDSE, if a transparent sheet of refractive index $\mu$ and thickness $t$ is introduced in front of one slit, then the optical path increases by
$$ (\mu - 1)t $$
Because of this, the central fringe shifts by a number of fringes
$$ n = \frac{(\mu - 1)t}{\lambda} $$
Here it is given that the central fringe shifts to the position of the $7^	ext{th}$ bright fringe.
So,
$$ \frac{(\mu - 1)t}{\lambda} = 7 $$
Given:
$$ \mu = 1.56,\qquad \lambda = 450\,	ext{nm} = 450 	imes 10^{-9}\,	ext{m} $$
Substitute:
$$ \frac{(1.56-1)t}{450	imes 10^{-9}} = 7 $$
$$ \frac{0.56\, t}{450	imes 10^{-9}} = 7 $$
So,
$$ t = \frac{7 	imes 450 	imes 10^{-9}}{0.56} $$
$$ t = \frac{3150 	imes 10^{-9}}{0.56} $$
$$ t = 5625 	imes 10^{-9}\,	ext{m} $$
Comparing with
$$ t = \alpha 	imes 10^{-9}\,	ext{m} $$
we get
$$ \alpha = 5625 $$
So, the answer is
$$ \boxed{5625} $$

Wave Optics

Integer / Numeric Answer Type