Question

In Young's experiment the monochromatic light is used to illuminate two slits A and B as shown in figure. Interference fringes are observed on a screen placed in front of the slits. Now a thin glass plate is placed normally in the path of the beam coming from the slit A, then.

A. there will be no change in fringe width.
B. fringe width will decrease
C. fringe width will increase
D. fringes will disappear

Answer 1

Hint: In order to solve this question, we will first draw the diagram by tracing the path of two light waves interfering due to slits at A and B at some point on screen and then calculate the total path difference of these two light rays and using general condition for bright fringe we will calculate fringe width.

Formula used:
If \[\Delta x\] is the net path difference and $n,\lambda $ is the number of bright fringe, wavelength of light used the, for bright fringe we have,
$\Delta x = n\lambda $
Fringe width is calculated by using
$\beta = {y_{n + 1}} - {y_n}$
where $y$ represents the distance of ${n^{th}}$ bright fringe from the center and $\beta $ denotes the fringe width.
Fringe width for Young’s double slit experiment is,
$\beta = \dfrac{{D\lambda }}{d}$

Complete step by step answer:
Let us first draw the diagram where slits are represented by ${S_1},{S_2}$ are at a distance of d and a screen is placed at a distance of D and two light waves passes through slits ${S_1},{S_2}$ and thin glass plate is placed in the path of wave going through ${S_1}$ Let, at point P the bright fringe is obtained.

So, if there were no glass plate then from young’s double slit experiment we know that the path difference is written as $\Delta x = {S_2}P - {S_1}P$ and for bright fringe forming at point P, it has a value of $\Delta x = {S_2}P - {S_1}P = \dfrac{{yd}}{D} \to (i)$

Now, when thing glass plate is placed then there will be extra path difference due to this glass plate along the path ${S_1}P$ and it will be equal to $\Delta {x_{plate}} = (\mu - 1)t$ so, new complete path difference can be written as $\Delta x' = {S_2}P - {S_1}P - \Delta {x_{plate}}$ and from equation (i) we get,
$\Delta x' = \dfrac{{yd}}{D} - (\mu - 1)t$
Now, for ${n^{th}}$ bright fringe formed at point P we have,
$\Delta x' = n\lambda $ on putting the values we get,
$\Rightarrow \dfrac{{{y_n}d}}{D} - (\mu - 1)t = n\lambda $
$\Rightarrow {y_n} = \dfrac{D}{d}[(\mu - 1)t + n\lambda ]$

Now, let us find the value of y for $n + 1$ so we have,
${y_{n + 1}} = \dfrac{D}{d}[(\mu - 1)t + (n + 1)\lambda ]$
For fringe width we have,
$\beta = {y_{n + 1}} - {y_n}$
On putting the values we get,
$\Rightarrow \beta = \dfrac{D}{d}[(\mu - 1)t + (n + 1)\lambda ] - [\dfrac{D}{d}[(\mu - 1)t + n\lambda ]]$
on solving we get,
$\beta = \dfrac{{D\lambda }}{d}$ and we also know that, in young’s double slit experiment the fringe width have a value of $\beta = \dfrac{{D\lambda }}{d}$ so, even after placing the glass plate the width of fringes don’t increase neither decrease.

Hence, the correct option is A.

Note: It should be remembered that, when a glass plate is placed on the path of wave the fringe width remain same but the position of bright fringes gets shifted in the direction where glass plate was placed and the amount to which position of fringes gets shifted is $\dfrac{D}{d}(\mu - 1)t$ and each fringes gets shifted to this amount which is the main reason of unchanged fringe width.