A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500 mathringA ?

Question

A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500 mathringA ? (A)  4400 mathringA (B)  4100 mathringA (C)  4642.8 mathringA  (D)  9100 mathringA

Tardigrade Admin · Accepted Answer

x=((2n+1)λ D/2a) For red light x=((4+1)D/2a)× 6500 For unknown wavelength of light, x=((6+1)D/2a)× λ Accordingly, ∴ 5× 6500=7× λ ⇒ λ =(5/7)× 6500 = text 4642.8 mathringA

Q. A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength $6500 \overset{˚}{A}$ ?

$4400 \overset{˚}{A}$

$4100 \overset{˚}{A}$

$4642.8 \overset{˚}{A}$

$9100 \overset{˚}{A}$

$x = \frac{( 2 n + 1 ) λ D}{2 a}$
For red light $x = \frac{( 4 + 1 ) D}{2 a} \times 6500$
For unknown wavelength of light,
$x = \frac{( 6 + 1 ) D}{2 a} \times λ$
Accordingly, $∴$ $5 \times 6500 = 7 \times λ$
$\Rightarrow$ $λ = \frac{5}{7} \times 6500$
$= 4642.8 \overset{˚}{A}$

Q. A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500A˚ ?

4400A˚

4100A˚

4642.8A˚

9100A˚