Question

What happens to the fringes pattern, if in the path of one of the slits a glass plate which absorbs $50\% $ energy is interposed?
(A) Brightness of fringes will decreases but the dark fringe will become brighter
(B) No fringes are observed
(C) The fringes width decreases
(D) None of the above

Answer 1

Hint: First we have to assume the intensity of two slits equals and if in the path of one slit a glass is placed then if the intensity gets reduced to one by two of its previous intensity. Then in this case the maximum intensity will decrease from its normal case and the minimum intensity will be greater than the normal case.

Complete step by step answer:
As per the problem we need to find what will happen to the fringes if the path of one of the slits to a glass plate which absorbs $50\% $ energy is interposed.
Now let this ace as a normal young’s double slit apparatus.
Where the intensity of both the slits equals.
$I_1 = I_2 = I_0$
Now on calculating the maximum intensity we will get,
$\operatorname{I} {\text{max}} = {\left( {\sqrt {I_1} + \sqrt {I_2} } \right)^2}$
Therefore the maximum intensity be,
$\operatorname{I} {\text{max = 4I_0}}$
Similarly the minimum intensity will be,
$\operatorname{I} {\text{min}} = {\left( {\sqrt {I_1} - \sqrt {I_2} } \right)^2}$
Therefore the maximum intensity be,
$\operatorname{I} {\text{min = 0}}$
When one slit is covered by a glass plate we will get,
$I_1 = \dfrac{{I_0}}{2}$
$I_2 = I_0$
In this if we find the maximum and minimum intensity then we can see that,
$\operatorname{I} {\text{max}}' < 4I_0$
And $\operatorname{I} {\text{min}}' > 0$
Therefore the correct option is (A) Brightness of fringes will decrease but the dark fringe will become brighter.

Note: We can also solve this problem by finding and comparing the path difference. We know that the path difference between the rays is zero then the center of maxima is formed. When a glass plate of some thickness let say t, is introduce than there is a change in path difference which can be represented as $\Delta x = \left( {\mu - 1} \right)t$ where $\mu \,and\,t$ are the refractive index of the plate and thickness of the plate respectively. So due to this the center of maxima will shift in the direction of the ray and this will cause a decrease in the brightness of the fringes.