In double slit experiment, the distance between two slits is 0.6 mm and these are illumunated with light of wavelength 4800 mathringA . The angular width of first dark fringe on the screen distant 120 cm from slits will be :

Question

In double slit experiment, the distance between two slits is 0.6 mm and these are illumunated with light of wavelength 4800 mathringA . The angular width of first dark fringe on the screen distant 120 cm from slits will be : (A)  8× 10-4rad  (B)  6× 10-4rad  (C)  4× 10-4rad  (D)  16× 10-4rad

Tardigrade Admin · Accepted Answer

Angular fringe width of first minima is (2x/D)=2(2n-1)(A/2d) =(2n-1)(λ /d) Given, d=0.6 text mm=0.6× 10-3m λ =4800AA=4.8× 10-7m, text n=1 ∴ (2x/D)=((2× 1-1)× 4.8× 10-7/0.6× 10-3) =8× 10-4rad.

Q. In double slit experiment, the distance between two slits is $0.6$ mm and these are illumunated with light of wavelength $4800 \overset{˚}{A}$ . The angular width of first dark fringe on the screen distant $120 c m$ from slits will be :

$8 \times 10^{- 4} r a d$

$6 \times 10^{- 4} r a d$

$4 \times 10^{- 4} r a d$

$16 \times 10^{- 4} r a d$

Q. In double slit experiment, the distance between two slits is 0.6 mm and these are illumunated with light of wavelength 4800A˚ . The angular width of first dark fringe on the screen distant 120cm from slits will be :

8×10−4rad

6×10−4rad

4×10−4rad

16×10−4rad

Angular fringe width of first minima is D2x​=2(2n−1)2dA​ =(2n−1)dλ​ Given, d=0.6mm=0.6×10−3m λ=4800AA=4.8×10−7m,n=1 ∴ D2x​=0.6×10−3(2×1−1)×4.8×10−7​ =8×10−4rad.

Q. In double slit experiment, the distance between two slits is $0.6$ mm and these are illumunated with light of wavelength $4800 \overset{˚}{A}$ . The angular width of first dark fringe on the screen distant $120 c m$ from slits will be :

$8 \times 10^{- 4} r a d$

$6 \times 10^{- 4} r a d$

$4 \times 10^{- 4} r a d$

$16 \times 10^{- 4} r a d$