Question

Light of wavelength $500{\rm{ nm}}$ is used to form interference patterns in Young’s double slit experiment. A uniform glass plate of refractive index $1.5$ and thickness $0.1{\rm{ mm}}$ is introduced in the path of one of the interfering beams. The number of fringes which will shift the cross wire due to this is:
(A) $400$
(B) $300$
(C) $200$
(D) $100$

Answer 1

Hint: The Young’s double slit experiment is used to show the dual nature of the light that the light can be defined as the waves and particles. In the experiment the interference of the light beam takes place, so for the interference in light having a wavelength $\lambda $ with a glass plate of refractive index $\mu $ and thickness $t$ the relationship is given by this expression-
$\left( {\mu - 1} \right)t = n\lambda $
Where, $n$ is the number of fringes.

Complete step by step answer:
Given:
The wavelength of the light $\lambda = 500{\rm{ nm}}$
Conversion formula - $1{\rm{ nanometer = 1}}{{\rm{0}}^{ - 9}}{\rm{ meter}}$
So, $\lambda = 500 \times {\rm{1}}{{\rm{0}}^{ - 9}}{\rm{ m}}$
The refractive index of the glass plate $\mu = 1.5$
And, the thickness of the glass plate $t = 0.1{\rm{ mm}}$
Conversion formula - $1{\rm{ millimeter = 1}}{{\rm{0}}^{ - 3}}{\rm{ meter}}$
So, $t = 0.1 \times {\rm{1}}{{\rm{0}}^{ - 3}}{\rm{ m}}$
Now using the formula, we have,
$\left( {\mu - 1} \right)t = n\lambda $
Substituting the values in the formula we get,
\[\begin{array}{l}
\left( {1.5 - 1} \right)0.1 \times {\rm{1}}{{\rm{0}}^{ - 3}} = n \times 500 \times {\rm{1}}{{\rm{0}}^{ - 9}}\\
n = \dfrac{{0.5 \times 0.1 \times {\rm{1}}{{\rm{0}}^{ - 3}}}}{{500 \times {\rm{1}}{{\rm{0}}^{ - 9}}}}
\end{array}\]
Solving this we get,
$n = 100$
Therefore, the number of fringes is $100$ and the correct option is (D) $100$.

Note: The number of fringes which will shift the cross wire when one of the slits is covered by a uniform glass plate is calculated in the Young’s Double Slit Experiment (YDSE). In this experiment, the entire interference pattern of the fringes would shift and the magnitude of this shift would be by a number of fringes. It should be noted that this shift is completely independent of the wavelength of the light.