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Asked by eknathmote50 | 24 May, 2019, 11:44: AM

Expert Answer

The intensity produced on the screen is directy proportional to the power coming out from the slit.

The power coming out of the slit is given by

P equals I w L w h e r e space I equals space i n t e n s i t y space o f space i n c i d e n t space b e a m w equals space w i d t h space o f space t h e space s l i t l equals l e n g t h space o f space t h e space s l i t

Hence power is directly proportionl to the width of the slit

As a result intensity produced on the screen

I subscript s

is directly proportional to the width

S subscript 1 space a n d space S subscript 2

are of equal width

Intensity produced will be

I subscript 1 equals I subscript 2 equals I subscript 0

Maximum and minimum intensity is given by

I subscript m a x end subscript equals open parentheses square root of I subscript 0 end root plus square root of I subscript 0 end root close parentheses squared equals 4 I subscript 0 I subscript m i n end subscript equals open parentheses square root of I subscript 0 end root minus square root of I subscript 0 end root close parentheses squared equals 0

Hence bright fringes and dark fringes are clearly distinguishable

If the slit width is not equal then then intensity produced by both the slits will not be equal

I subscript 1 not equal to I subscript 2 a n d space h e n c e I subscript m a x end subscript equals open parentheses square root of I subscript 1 end root plus square root of I subscript 2 end root close parentheses squared not equal to 4 I subscript 0 I subscript m i n end subscript equals open parentheses open parentheses square root of I subscript 1 end root minus square root of I subscript 2 end root close parentheses squared close parentheses squared not equal to 0

Since the minimum intensity is not equal to zero there won't be any dark spot on the screen.

Answered by Utkarsh Lokhande | 25 May, 2019, 07:36: PM

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