Light of wavelength λ=5000 mathringA falls normally on a narrow slit. A screen is placed at a distance of 1 m from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5 mm from the centre of central maximum. The width of the slit is

Question

Light of wavelength λ=5000 mathringA falls normally on a narrow slit. A screen is placed at a distance of 1 m from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5 mm from the centre of central maximum. The width of the slit is (A) 0.1 mm (B) 1.0 mm (C) 0.5 mm (D) 0.2 mm

Tardigrade Admin · Accepted Answer

Position of n th minima xn=(n λ D/d) ⇒ 5 × 10-3=(1 × 5000 × 10-10 × 1/d) ⇒ d=10-4 m =0.1 mm.

Q. Light of wavelength λ=5000A˚ falls normally on a narrow slit. A screen is placed at a distance of 1m from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5mm from the centre of central maximum. The width of the slit is

0.1 mm

1.0 mm

0.5 mm

0.2 mm

Position of n th minima xn​=dnλD​ ⇒5×10−3=d1×5000×10−10×1​ ⇒d=10−4m=0.1mm.

Position of $n$ th minima $x_{n} = \frac{nλ D}{d}$
$\Rightarrow 5 \times 1 0^{- 3} = \frac{1 \times 5000 \times 1 0 ^{- 10} \times 1}{d}$
$\Rightarrow d = 1 0^{- 4} m = 0.1 mm$ .