In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500 mathringA , the j first minimum is observed, at angle 30°. The first secondary maximum is observed at an angle θ =

Question

In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500 mathringA , the j first minimum is observed, at angle 30°. The first secondary maximum is observed at an angle θ = (A) sin-1((1/√2)) (B) sin-1((1/4)) (C) sin-1((3/4)) (D) sin-1((√3/2))

Tardigrade Admin · Accepted Answer

Slit width = d λ = 5500 mathringA = 5.5 × 10-7m, θn = 30° For first secondary minima, d sin θn = λ d =(λ/sin θn) = (5.5 × 10-7 /sin 30°) = 11× 10-7 m For first secondary maxima, d sin θn = (3λ/2) i.e., sin θn = (3λ/2d) =(3× 5.5× 10-7/2× 11× 10-7) sin θn = (3/4 ) or θn = sin -1 (3/4)

Q. In a Fraunhofer diffraction at single slit of width $d$ with incident light of wavelength $5500 \overset{˚}{A}$ , the j first minimum is observed, at angle $3 0^{\circ}$ . The first secondary maximum is observed at an angle $θ =$

$s i n^{- 1} (\frac{1}{2})$

$s i n^{- 1} (\frac{1}{4})$

$s i n^{- 1} (\frac{3}{4})$

$s i n^{- 1} (\frac{3}{2})$

Q. In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500A˚, the j first minimum is observed, at angle 30∘. The first secondary maximum is observed at an angle θ=

sin−1(2​1​)

sin−1(41​)

sin−1(43​)

sin−1(23​​)

Slit width =d λ=5500A˚=5.5×10−7m,θn​=30∘ For first secondary minima, dsinθn​=λ d=sinθn​λ​=sin30∘5.5×10−7​ =11×10−7m For first secondary maxima, dsinθn​=23λ​ i.e., sinθn​=2d3λ​ =2×11×10−73×5.5×10−7​ sinθn​=43​ or θn​=sin−1(3/4)

Q. In a Fraunhofer diffraction at single slit of width $d$ with incident light of wavelength $5500 \overset{˚}{A}$ , the j first minimum is observed, at angle $3 0^{\circ}$ . The first secondary maximum is observed at an angle $θ =$

$s i n^{- 1} (\frac{1}{2})$

$s i n^{- 1} (\frac{1}{4})$

$s i n^{- 1} (\frac{3}{4})$

$s i n^{- 1} (\frac{3}{2})$