
What is the effect on the fringe width of interference fringes in a young’s double slit experiment due to each of the following operations?
$(a)$ The screen is moved away from the plane of the slits.
$(b)$ The (monochromatic) source is replaced by another (monochromatic) source of the shorter wavelength
$(c)$ The separation between the two slits is increased
$(d)$ The width of two slits are slightly increased (In each operation, take all parameters, other than one specified to remain unchanged)
Answer
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Hint: In this question use the direct formula for Band width of young’s double slit experiment that is $B.W = \dfrac{D}{d}\lambda $, where D is the distance between the screen and the slits, d is the distance the slits, $\lambda $ is the wavelength. Variations of these parameters have direct influence over band width. This will help getting the answer.
Formula used – $B.W = \dfrac{D}{d}\lambda $
Complete Step-by-Step solution:
As we know in a Young’s double slit experiment the bandwidth (B.W) is calculated as:
$B.W = \dfrac{D}{d}\lambda $................. (1), where D = distance between the screen and the slits.
d = distance between the slits.
$\lambda $ = wavelength.
$\left( a \right)$ When the screen is moved away from the plane of the slits the distance between the slits is increased from equation (1).
Therefore if D is increased, and it is given that remaining parameters remain same so bandwidth (B.W) is increased.
$\left( b \right)$ When the monochromatic source is replaced by another monochromatic source of shorter wavelength, hence the bandwidth (B.W) is decreased as wavelength $\left( \lambda \right)$ is decreased from equation (1).
$\left( c \right)$ When the separation between the two slits is increased so the distance (d) between the slits is increased, hence the bandwidth (B.W) is decreased from equation (1).
$\left( d \right)$ If we increase the slit width so the separation between the slits also increases hence d is increased, so if d is increased bandwidth (B.W) is decreased from equation (1).
So this is the required answer.
Note – When monochromatic light passing through two narrow slits illuminates a distant screen, a characteristic pattern of bright and dark fringes is observed. This interference pattern is caused by the superposition of overlapping light waves originating from the two slits.
Formula used – $B.W = \dfrac{D}{d}\lambda $
Complete Step-by-Step solution:
As we know in a Young’s double slit experiment the bandwidth (B.W) is calculated as:
$B.W = \dfrac{D}{d}\lambda $................. (1), where D = distance between the screen and the slits.
d = distance between the slits.
$\lambda $ = wavelength.
$\left( a \right)$ When the screen is moved away from the plane of the slits the distance between the slits is increased from equation (1).
Therefore if D is increased, and it is given that remaining parameters remain same so bandwidth (B.W) is increased.
$\left( b \right)$ When the monochromatic source is replaced by another monochromatic source of shorter wavelength, hence the bandwidth (B.W) is decreased as wavelength $\left( \lambda \right)$ is decreased from equation (1).
$\left( c \right)$ When the separation between the two slits is increased so the distance (d) between the slits is increased, hence the bandwidth (B.W) is decreased from equation (1).
$\left( d \right)$ If we increase the slit width so the separation between the slits also increases hence d is increased, so if d is increased bandwidth (B.W) is decreased from equation (1).
So this is the required answer.
Note – When monochromatic light passing through two narrow slits illuminates a distant screen, a characteristic pattern of bright and dark fringes is observed. This interference pattern is caused by the superposition of overlapping light waves originating from the two slits.
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