In a Young's double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle $\frac{1}{40} \mathrm{rad}$ by using light of wavelength $\lambda_{1}$. When the light of wavelength $\lambda_{2}$ is used a bright fringe is seen at the same angle in the same set up. Given that $\lambda_{1}$ and $\lambda_{2}$ are in visible range (380 nm to 740 nm), their values are:

(A) $625 \mathrm{~nm}, 500 \mathrm{~nm}$

(B) $380 \mathrm{~nm}, 525 \mathrm{~nm}$

(C) $380 \mathrm{~nm}, 500 \mathrm{~nm}$

(D) $400 \mathrm{~nm}, 500 \mathrm{~nm}$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Wave Optics

In a Young's double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle

\frac{1}{40} rad

by using light of wavelength

λ_{1}

. When the light of wavelength

λ_{2}

is used a bright fringe is seen at the same angle in the same set up. Given that

λ_{1}

and

λ_{2}

are in visible range (380 nm to 740 nm), their values are:

Answer: (A)

A bright fringe sits where the path difference is a whole number of wavelengths,

d sin θ = nλ

, and for the small given angle

sin θ \approx θ = \frac{1}{40}

. So

λ = \frac{d θ}{n} = \frac{0.1 \times 1 0 ^{- 3} \times ( 1/40 )}{n} = \frac{2500}{n}

nm. Only orders that land the wavelength inside

380

740

nm are allowed:

n = 4

gives

625

nm and

n = 5

gives

500

nm, while neighbouring orders fall outside the visible band. The same direction can thus be bright for two colours at two different orders,

625

nm and

500

nm.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option