
In Young’s experiment, monochromatic light through a single slit S is used to illuminate the two slits ${{S}_{1}}$and ${{S}_{2}}$. Interference fringes are obtained on a screen. The fringe width is found to be $w$. Now a thin sheet of mica (thickness $t$ and refractive index $\mu $) is placed near and in front of one of the two slits. Now the fringe width is found to be $w'$, then
(A) $w'={}^{w}/{}_{\mu }$
(B) $w'=w\mu $
(C) $w'=(\mu -1)tw$
(D) $w'=w$
Answer
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Hint: Young’s experiment is a classical investigation into the nature of light, an investigation that provided the basic element in the development of the wave theory. In this experiment, Young identified the phenomenon called interference. Observing that when light from a single source is split into two beams, and the two beams are then recombined, the combined beam shows a pattern of light and dark fringes, Young concluded that the fringes result from the fact that when the beams recombine their peaks and troughs may not be in phase.
Complete step by step answer:
Given:
Fringe width $=w$
Thickness $=t$
Refractive index $=\mu $
We know that the fringe of width can be defined as:
$w=\dfrac{D\lambda }{d}\,\,\,\,.....(1)$
The distance between two consecutive bright or dark fringes is called the fringe width.
The distance of fringes when a thin sheet of mica is placed near and in front of one of the two slits is given by:
${{y}_{n}}=\dfrac{D}{d}(n\lambda +(\mu -1)t)\,\,\,\,.....(2)$
To calculate the width of fringe, we will use its formula which is given by:
$w'={{y}_{n+1}}-{{y}_{n}}\,\,\,\,.....(3)$
Using equation (1) and (2), we get:
$w'=\dfrac{D}{d}((n+1)\lambda +(\mu -1)t)-\dfrac{D}{d}(n\lambda +(\mu -1)t)$
By solving the above equation we get:
$w'=\dfrac{D\lambda }{d}\,\,\,\,.....(4)$
From equation (1) and (2), we get:
$w'=w$
Therefore, the fringe width remains the same. Hence option D. is correct.
Note: The interference pattern obtained in the double-slit experiment consists of alternate bright and dark fringes that are parallel to the slits. The fringe formed at the center of the fringe pattern is called the central bright fringe. To solve this kind of question, we should know about interference and young people's experiments.
Complete step by step answer:
Given:
Fringe width $=w$
Thickness $=t$
Refractive index $=\mu $
We know that the fringe of width can be defined as:
$w=\dfrac{D\lambda }{d}\,\,\,\,.....(1)$
The distance between two consecutive bright or dark fringes is called the fringe width.
The distance of fringes when a thin sheet of mica is placed near and in front of one of the two slits is given by:
${{y}_{n}}=\dfrac{D}{d}(n\lambda +(\mu -1)t)\,\,\,\,.....(2)$
To calculate the width of fringe, we will use its formula which is given by:
$w'={{y}_{n+1}}-{{y}_{n}}\,\,\,\,.....(3)$
Using equation (1) and (2), we get:
$w'=\dfrac{D}{d}((n+1)\lambda +(\mu -1)t)-\dfrac{D}{d}(n\lambda +(\mu -1)t)$
By solving the above equation we get:
$w'=\dfrac{D\lambda }{d}\,\,\,\,.....(4)$
From equation (1) and (2), we get:
$w'=w$
Therefore, the fringe width remains the same. Hence option D. is correct.
Note: The interference pattern obtained in the double-slit experiment consists of alternate bright and dark fringes that are parallel to the slits. The fringe formed at the center of the fringe pattern is called the central bright fringe. To solve this kind of question, we should know about interference and young people's experiments.
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