Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm . The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

Question

Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm . The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum) (A) 6× 10- 3 rad (B) 4× 10- 3 rad (C) 2.4× 10- 3 rad (D) 4.5× 10- 3 rad

Tardigrade Admin · Accepted Answer

λ =600 nm=600× 10- 9m d=0.2 mm=0.2× 10- 3m <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-vm55xarez0oo.png" /> ∵ Linear width of central maximum, x=(2 θ )D Where, 2θ: Angular width of central maximum. We know that, x=2(λ D/d) Angular width 2θ .D=2(λ D/d) ⇒ 2θ =(2 λ /d)=(2 × 600 × 1 0- 9/0.2 × 1 0- 3) =6× 10- 7× 104 =6× 10- 3 rad

Q. Light of wavelength $600 nm$ is incident normally on a slit of width $0.2 mm$ . The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

$6 \times 1 0^{- 3} r a d$

$4 \times 1 0^{- 3} r a d$

$2.4 \times 1 0^{- 3} r a d$

$4.5 \times 1 0^{- 3} r a d$

Q. Light of wavelength 600nm is incident normally on a slit of width 0.2mm . The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

6×10−3rad

4×10−3rad

2.4×10−3rad

4.5×10−3rad

λ=600nm=600×10−9m d=0.2mm=0.2×10−3m ∵ Linear width of central maximum, x=(2θ)D Where, 2θ: Angular width of central maximum. We know that, x=2dλD​ Angular width 2θ.D=2dλD​ ⇒2θ=d2λ​=0.2×10−32×600×10−9​ =6×10−7×104 =6×10−3rad

Q. Light of wavelength $600 nm$ is incident normally on a slit of width $0.2 mm$ . The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

$6 \times 1 0^{- 3} r a d$

$4 \times 1 0^{- 3} r a d$

$2.4 \times 1 0^{- 3} r a d$

$4.5 \times 1 0^{- 3} r a d$