(C) In Young's double slit experiment,the position of the $n^{th}$ bright fringe is given by $y_n = n \frac{\lambda D}{d}$.For the first case,$n_1 = 5

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(C) In Young's double slit experiment,the position of the $n^{th}$ bright fringe is given by $y_n = n \frac{\lambda D}{d}$.For the first case,$n_1 = 5$,$\lambda_1 = 600 nm$,and $y_1 = 6 mm$.So,$6 = 5 \times \frac{600 D}{d} \Rightarrow \frac{D}{d} = \frac{6}{5 \times 600} = \frac{1}{500} mm/nm$.For the second case,$n_2 = 3$,$\lambda_2 = 400 nm$.The position of the third bright fringe is $y_2 = n_2 \frac{\lambda_2 D}{d}$.Substituting the values: $y_2 = 3 \times 400 \times \frac{1}{500} = \frac{1200}{500} = 2.4 mm$.

Class 12 Physics Wave Optics Young's Double Slit Experiment (YDSE)

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When monochromatic light of wavelength $600 nm$ is used in Young's double slit experiment,the fifth order bright fringe is formed at $6 mm$ from the central bright fringe on the screen. If the experiment is conducted with light of wavelength $400 nm$,the third order bright fringe will be located at: (in $mm$)

AP EAMCET 2022 All AP EAMCET

A
$1.6$
B
$2$
C
$2.4$
D
$3$

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Class 12

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Wave Optics

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Young's Double Slit Experiment (YDSE)

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