The fringe width at a distance of 50 cm from the slits in Young's experiment for light of wavelength 6000 mathringA is 0.048 cm. The fringe width at the same distance for λ=5000 mathringA, will be

Question

The fringe width at a distance of 50 cm from the slits in Young's experiment for light of wavelength 6000 mathringA is 0.048 cm. The fringe width at the same distance for λ=5000 mathringA, will be (A) 0.04 cm (B) 0.4 cm (C) 0.14 cm (D) 0.45 cm

Tardigrade Admin · Accepted Answer

As, β'=(λ'/λ) β=(5000/6000) × 0.48=0.04 cm

Q. The fringe width at a distance of $50 c m$ from the slits in Young's experiment for light of wavelength $6000 \overset{˚}{A}$ is $0.048 c m$ . The fringe width at the same distance for $λ = 5000 \overset{˚}{A}$ , will be

0.04 cm

0.4 cm

0.14 cm

0.45 cm

As, $β^{'} = \frac{λ ^{'}}{λ} β = \frac{5000}{6000} \times 0.48 = 0.04 c m$

Q. The fringe width at a distance of 50cm from the slits in Young's experiment for light of wavelength 6000A˚ is 0.048cm. The fringe width at the same distance for λ=5000A˚, will be