A flake of glass (refractive index mu = 1.5) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The fringe shift $\Delta x$ due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{D}{

Vedclass · Accepted Answer

$8400$

Vedclass · Answer

(C) The fringe shift $\Delta x$ due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{D}{	ext{d}}(\mu - 1)t$.Since the fringe width is $\beta = \frac{\lambda D}{	ext{d}}$,we can write the shift in terms of the number of fringes $n$ as $\Delta x = n\beta$.Equating the two expressions: $n\beta = \frac{D}{	ext{d}}(\mu - 1)t$.Substituting $\beta = \frac{\lambda D}{	ext{d}}$,we get $n \left( \frac{\lambda D}{	ext{d}} ight) = \frac{D}{	ext{d}}(\mu - 1)t$.This simplifies to $n\lambda = (\mu - 1)t$.Given $n = 7$,$\lambda = 600 \, nm$,and $\mu = 1.5$,we have $7 	imes 600 = (1.5 - 1)t$.$4200 = 0.5 	imes t$.$t = \frac{4200}{0.5} = 8400 \, nm$.

Similar Questions

$A$ double slit experiment is performed with light of wavelength $500\, nm$. $A$ thin film of thickness $2\, \mu m$ and refractive index $1.5$ is introduced in the path of the upper beam. The location of the central maximum will

$A$ transparent film $(\mu=1.45)$ of thickness $0.02 \ mm$ is placed on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength $620 \ nm$. How many fringes will cross through the center if the film is removed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module