In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength of light used is 5000 mathring A then the distance between the first minimum on either side of the central maximum is (θ is small and measured in radian)

Question

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength of light used is 5000 mathring A then the distance between the first minimum on either side of the central maximum is (θ is small and measured in radian) (A) 10-1 m  (B) 10-2 m  (C)  2 × 10-2 m  (D)  2 × 10-1 m

Tardigrade Admin · Accepted Answer

Distance between the first 2 minimum on other side =(2 λ D / d ) =(2 × 5 × 103 × 10-10 × 2/2 × 10-1 × 10-3) =10 × (10-7/10-4) =10-2 m

Q. In Fraunhofer diffraction pattern, slit width is $0.2 mm$ and screen is at $2 m$ away from the lens. If wavelength of light used is $5000 \overset{˚}{A}$ then the distance between the first minimum on either side of the central maximum is ( $θ$ is small and measured in radian)

$1 0^{- 1} m$

$1 0^{- 2} m$

$2 \times 1 0^{- 2} m$

$2 \times 1 0^{- 1} m$

Distance between the first $2$ minimum on other side $= \frac{2 λ D}{d}$
$= \frac{2 \times 5 \times 1 0 ^{3} \times 1 0 ^{- 10} \times 2}{2 \times 1 0 ^{- 1} \times 1 0 ^{- 3}}$
$= 10 \times \frac{1 0 ^{- 7}}{1 0 ^{- 4}}$
$= 1 0^{- 2} m$

Q. In Fraunhofer diffraction pattern, slit width is 0.2mm and screen is at 2m away from the lens. If wavelength of light used is 5000A˚ then the distance between the first minimum on either side of the central maximum is (θ is small and measured in radian)

10−1m

10−2m

2×10−2m

2×10−1m