Question

The index of refraction of a glass plate is 1.48 at ${\theta _1} = 30^\circ C$ and varies linearly with temperature with a coefficient of $2 \cdot 5 \times {10^{ - 5}}^\circ {C^{ - 1}}$. The coefficient of linear expansion of the glass is $0 \cdot 5 \times {10^{ - 5}}^\circ {C^{ - 1}}$. At $30^\circ C$, the length of the glass plate is 3cm. This plate is placed in front of one the slits in Young’s double-slit temperature increases at a rate of $\dfrac{{5^\circ C}}{{\min .}}$, the light source has wavelength $\lambda = 589nm$ and the glass plate initially is at $\theta = 30^\circ C$. The number of fringes that shift on the screen in each minute is nearly (use approximately):
A) $1$
B) $11$
C) $110$
D) $1 \cdot 1 \times {10^3}$.

Answer 1

Hint: The path difference is the difference between the path of the actual and final wave after getting pass from the slab.

Formula used:
The formula of the path difference is given by,
$ \Rightarrow P \cdot D = \mu - {\mu _o}{t_o} = n{\lambda _o}$
Where the path difference is P.D, the refractive index is $\mu $ and ${\mu _o}$ at different points of time, the number of fringes is n and the wavelength is${\lambda _o}$.

Complete step by step solution:
It is given in the problem that the index of refraction of a glass plate is 1.48 at ${\theta _1} = 30^\circ C$ and varies linearly with temperature with a coefficient of $2 \cdot 5 \times {10^{ - 5}}^\circ {C^{ - 1}}$, the coefficient of linear expansion of the glass is $0 \cdot 5 \times {10^{ - 5}}^\circ {C^{ - 1}}$ at $30^\circ C$, the length of the glass plate is 3cm the plate is placed in front of one the slits in Young’s double-slit temperature increases at a rate of $\dfrac{{5^\circ C}}{{\min .}}$, the light source has wavelength $\lambda = 589nm$ and the glass plate initially is at $\theta = 30^\circ C$ then we need to find the number of fringes shit on the screen in each minute.
The path difference is given by,
$ \Rightarrow P \cdot D = \mu - {\mu _o}{t_o} = n{\lambda _o}$
Where the path difference is P.D, the refractive index is $\mu $ and ${\mu _o}$ at different points of time, the number of fringes is n and the wavelength is${\lambda _o}$.
Also the refractive index can be written as,
$ \Rightarrow \mu = {\mu _o}\left( {1 + \alpha \theta } \right){t_o}\left( {1 + \beta \theta } \right)$
Therefore we get,
$n\lambda = {t_o}\left[ {\mu \left( {\alpha \theta + \beta \theta + \alpha \beta {\theta ^2}} \right) - \alpha \theta } \right]$
$ \Rightarrow n = \dfrac{{{t_o}\left[ {\mu \left( {\alpha \theta + \beta \theta + \alpha \beta {\theta ^2}} \right) - \alpha \theta } \right]}}{\lambda }$
$ \Rightarrow n = \dfrac{{{\mu _o}{t_o}\left( {\alpha + \beta } \right)\theta }}{{{\lambda _o}}}$
As the given values are ${\mu _o} = 1 \cdot 48$, ${t_o} = 3 \times {10^{ - 2}}m$, $\alpha = 2 \cdot 5 \times {10^{ - 5}}^\circ {C^{ - 1}}$, $\beta = 0 \cdot 5 \times {10^{ - 5}}^\circ {C^{ - 1}}$, $\theta = 5^\circ C\min {.^{ - 1}}$ and the wavelength is equal to ${\lambda _o} = 589 \times {10^{ - 9}}m$. Putting these values in the above relation we get,
$ \Rightarrow n = \dfrac{{{\mu _o}{t_o}\left( {\alpha + \beta } \right)\theta }}{{{\lambda _o}}}$
$ \Rightarrow n = \dfrac{{1 \cdot 48 \times 3 \times {{10}^{ - 2}}\left( {2 \cdot 5 \times {{10}^{ - 5}} + 0 \cdot 5 \times {{10}^{ - 5}}} \right) \times 5}}{{589 \times {{10}^{ - 9}}}}$
$ \Rightarrow n = \dfrac{{1 \cdot 48 \times 3 \times {{10}^{ - 2}}\left( {2 \cdot 5 \times {{10}^{ - 5}} + 0 \cdot 5 \times {{10}^{ - 5}}} \right) \times 5}}{{589 \times {{10}^{ - 9}}}}$
$ \Rightarrow n = \dfrac{{1 \cdot 48 \times 3 \times {{10}^{ - 2}} \times 3 \times {{10}^{ - 5}} \times 5}}{{589 \times {{10}^{ - 9}}}}$
$ \Rightarrow n = \dfrac{{66 \cdot 6 \times {{10}^{ - 7}}}}{{589 \times {{10}^{ - 9}}}}$
$ \Rightarrow n = \dfrac{{66 \cdot 6 \times {{10}^2}}}{{589}}$
$ \Rightarrow n = \dfrac{{6660}}{{589}}$
$ \Rightarrow n = 11 \cdot 3$
$ \Rightarrow n \approx 11$.
The number of fringes that shift on the screen in each minute is equal $n = 11$.

The correct answer for this problem is option B.

Note: The refractive index of the glass is changing linearly with the temperature. If the beam of ray passes through the slab and then comes out of the slab then there is a difference in between the initial path and the final path it is known as path difference.