In a Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of 30° Then the direction θ of the first secondary maximum is

Question

In a Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of 30° Then the direction θ of the first secondary maximum is (A)  tan -1((4/3)) (B) 60° (C)  sin -1((3/4)) (D)  tan -1((3/4))

Tardigrade Admin · Accepted Answer

For first minima sin θ=( n /d) ⇒ sin 30°=(400 × 10-9 m /d) or d=(400 × 10-9 m /(0.5))=800 × 10-9 m. Again sin θ=(3 λ/2 d)=(3/2) ((400 × 10-9/800 × 10-9))=(3/4) θ= sin -1((3/4))

Q. In a Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of $3 0^{\circ}$ Then the direction $θ$ of the first secondary maximum is

$tan^{- 1} (\frac{4}{3})$

$6 0^{\circ}$

$sin^{- 1} (\frac{3}{4})$

$tan^{- 1} (\frac{3}{4})$

Q. In a Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of 30∘ Then the direction θ of the first secondary maximum is

tan−1(34​)

60∘

sin−1(43​)

tan−1(43​)

For first minima sinθ=dn​ ⇒sin30∘=d400×10−9m​ or d=(0.5)400×10−9m​=800×10−9m. Again sinθ=2d3λ​=23​ (800×10−9400×10−9​)=43​ θ=sin−1(43​)

Q. In a Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of $3 0^{\circ}$ Then the direction $θ$ of the first secondary maximum is

$tan^{- 1} (\frac{4}{3})$

$6 0^{\circ}$

$sin^{- 1} (\frac{3}{4})$

$tan^{- 1} (\frac{3}{4})$