Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is 312.5× 10- 3 cm . If waves were incident normally on the slit, then wavelength of the light from the source will be

Question

Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is 312.5× 10- 3 cm . If waves were incident normally on the slit, then wavelength of the light from the source will be (A) 6000 Å (B) 6250 Å (C) 6400 Å (D) 6525 Å

Tardigrade Admin · Accepted Answer

Width of central max: = (2 λ D/a) λ = (0.2 × 10- 2 × 312.5 × 10- 3 × 10- 2 m/2 × (1)2) = 6250 × 10- 10 m = 6250 â«

Q. Width of the principal maximum on a screen at a distance of $50 c m$ from the slit having width $0.02 c m$ is $312.5 \times 1 0^{- 3} c m$ . If waves were incident normally on the slit, then wavelength of the light from the source will be

$6000 Å$

$6250 Å$

$6400 Å$

$6525 Å$

Width of central max: $= \frac{2 λ D}{a}$
$λ = \frac{0.2 \times 1 0 ^{- 2} \times 312.5 \times 1 0 ^{- 3} \times 1 0 ^{- 2} m}{2 \times \frac{1}{2}}$
$= 6250 \times 1 0^{- 10} m = 6250 Å$

Q. Width of the principal maximum on a screen at a distance of 50cm from the slit having width 0.02cm is 312.5×10−3cm . If waves were incident normally on the slit, then wavelength of the light from the source will be

6000Å

6250Å

6400Å

6525Å