The angular width of the central maximum in a single slit diffraction pattern is $60 ° .$ The width of the slit is $1 μ m .$ The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1cm, what is slit separation distance?
(i.e. distance between the centre of each slit)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

50 $μ m$

100 $μ m$

75 $μ m$

25 $μ m$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Angular position of first minima
$b \sin θ = λ$ $\Rightarrow \sin θ = \frac{λ}{b}$

So wavelength of light used
$λ = b \sin λ = b \sin 30 ° = \frac{b}{2}$
Given $2 θ = 60 °$
$\Rightarrow θ = 30 °$
If similar two slits are used in YDSE at a separation ‘d’ then we have
Question Image
Fringe width in YDSE interference pattern is given as
$β = \frac{λ D}{d} = \frac{b D}{2 d}$

$So, slit separation is$

$d = \frac{b D}{2 β} = \frac{10^{- 6} \times 0.5}{2 \times 0.01} = 25 \times 10^{- 6} m$ $\Rightarrow d = 25 μ m$