Question

Q.14.In Young's double slit experiment, the slit
separation is reduced to half and distance of the
screen is increased by 50 %. If X is the initial
fringe width then new fringe width is
O a) X/4
O b) 3X/4
O c) 3X / 2
O d) 3X​

Answer 1

Solution :-

we know that, distance between two adjacent bright fringes is called the fringe width . It is given by :-

• X = λD/d.

given that, the slit separation is reduced to half .

so,

→ New slit separation distance(d') = Half of d = (d/2)

and,

→ New screen distance(D') = 50% increased = 150% of D = (150*D)/100 = (3D/2)

then,

→ New Fringe(X') = λD'/d'

→ X' = λ(3D/2)/(d/2)

→ X' = λ(3D * 2) / (2d)

→ X' = λ(3D/d)

→ X' = 3[λD/d]

→ X' = 3X (d) (Ans.)

Hence, new fringe width is 3X .

Answer 2

General Expression of Fringe Width in YDSE experiment is :

$\beta = \dfrac{ \lambda D}{d}$

• $\lambda$ is wavelength of light, D is distance of screen and 'd' is slit separation.

Now, new values :

$D_{2} = D \times \dfrac{150}{100} = \dfrac{3D}{2}$

$d_{2} = \dfrac{d}{2}$

Now, new fringe width is :

$\beta_{2} = \dfrac{ \lambda D_{2} }{d_{2} }$

$\implies \beta_{2} = \dfrac{ \lambda (\dfrac{3D}{2} )}{ (\dfrac{d}{2} )}$

$\implies \beta_{2} = 3 \times \dfrac{ \lambda D}{ d}$

$\implies \beta_{2} = 3 \times \beta$

$\implies \beta_{2} = 3 X$

So, new fringe width is 3X.