In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the

Correct answer: B. 1 .9 mm Angular fringe width, θ = (λ)/(d), θ _ 1 θ _ 2 = d _ 2 d _ 1 (0.20)/(0.21) = d _ 2 2 ⇒ d _ 2 = (0.20)/(0.21) × 2 = 1.9mm

NEET 2018

NEET 2018 · Physics · Wave Optics

In Young's double slit experiment the separation d between the...

In Young's double slit experiment the separation

d

between the slits is

2 mm,

the wavelength

\lambda

of the light used is

5896Å

and distance

D

between the screen and slits is

100 cm.

It is found that the angular width of the fringes is

0.20 °

. To increase the fringe angular width to

{0.21}^{o}

(with same

\lambda\text{and}D

) the separation between the slits needs to be changed to

B. $1 .9 mm$

Angular fringe width,

\theta = \dfrac{\lambda}{d}, \dfrac{{\theta}_{1}}{{\theta}_{2}} = \dfrac{{d}_{2}}{{d}_{1}}

\dfrac{0.20}{0.21} = \dfrac{{d}_{2}}{2} ⇒ {d}_{2} = \dfrac{0.20}{0.21} \times 2 = 1.9mm