In a Young's double slit experiment, the separation d between the slits is 2 mm, the wavelength $λ$ of the light used is 5896 $\overset{0}{A}$ and distance D between the screen and slit is 100 cm. It is found that, the angular width of the fringe is 0.20⁰. To increase the fringe angular width to 0.21⁰ (with same $λ$ and D), the separation between the slits needs to be changed
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1.8 mm

1.7 mm

2.1 mm

1.9 mm

answer is B.

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Detailed Solution

In a Young's double slit experiment, angular width of a fringe is given as

$θ = \frac{λ}{d}$

where, $λ$ is the wavelength of the light source and d is the distance between the two slits.

$\Rightarrow θ \propto \frac{1}{d} or \frac{θ_{1}}{θ_{2}} = \frac{d_{2}}{d_{1}}$ …(i)

Given, $θ_{1} = 0.20 °, θ_{2} = 0.21 °$

d₁ = 2 mm

Substituting the given values in Eq. (i), we get

$\frac{0.20 °}{0.21 °} = \frac{d_{2}}{2}$

$\Rightarrow d_{2} = 2 \times \frac{0.20 °}{0.21 °}$

$= \frac{0.40}{0.21} = 1.9 mm$

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