1. In Young's double slit experiment, two slits are 1 mm apart, and the screen is placed 1 m away from the slits. Calculate the fringe width when light of wavelength 500 nm is used. 2. What should be the width of each slit in order to obtain 10 maxima of the double slits pattern within the central maximum of the single slit pattern?

1. $\beta = \frac{\lambda\text{D}}{\text{d}}$ $ = \frac{500\times10^{-9}\times1}{10^{-3}}$ $= 0.5\ mm or 5 x 10^{-4}m$ 2. $\beta_{0} = \frac{2\lambda\text{D}}{\text{a}} = 10 \beta$ $\text{a} = \frac{2\times500\times10^{-9}\times1}{10\times5\times10^{-4}}$ $? =2 \times 10^{−4}m\ or\ 0.2\ mm.$

1. In Young's double slit experiment, two slits are 1 mm apart, and the screen is placed 1 m away from the slits. Calculate the fringe width when light of wavelength 500 nm is used. 2. What should be the width of each slit in order to obtain 10 maxima of the double slits pattern within the central maximum of the single slit pattern?

1. $\beta = \frac{\lambda\text{D}}{\text{d}}$ $ = \frac{500\times10^{-9}\times1}{10^{-3}}$ $= 0.5\ mm or 5 x 10^{-4}m$ 2. $\beta_{0} = \frac{2\lambda\text{D}}{\text{a}} = 10 \beta$ $\text{a} = \frac{2\times500\times10^{-9}\times1}{10\times5\times10^{-4}}$ $? =2 \times 10^{−4}m\ or\ 0.2\ mm.$

1. In Young's double slit experiment, two slits are 1 mm apart, and the screen is placed 1 m away from the slits. Calculate the fringe width when light of wavelength 500 nm is used. 2. What should be the width of each slit in order to obtain 10 maxima of the double slits pattern within the central maximum of the single slit pattern?

1. $\beta = \frac{\lambda\text{D}}{\text{d}}$ $ = \frac{500\times10^{-9}\times1}{10^{-3}}$ $= 0.5\ mm or 5 x 10^{-4}m$ 2. $\beta_{0} = \frac{2\lambda\text{D}}{\text{a}} = 10 \beta$ $\text{a} = \frac{2\times500\times10^{-9}\times1}{10\times5\times10^{-4}}$ $? =2 \times 10^{−4}m\ or\ 0.2\ mm.$

1. In Young's double slit experiment, two slits are 1 mm apart, and the screen is placed 1 m away from the slits. Calculate

3 Marks Question

GSEB - English Medium

In Young's double slit experiment, two slits are 1 mm apart, and the screen is placed 1 m away from the slits. Calculate the fringe width when light of wavelength 500 nm is used.
What should be the width of each slit in order to obtain 10 maxima of the double slits pattern within the central maximum of the single slit pattern?

CBSE OUTSIDE DELHI - SET 2 EAST 2016

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Solution

$\beta = \frac{\lambda\text{D}}{\text{d}}$

$ = \frac{500\times10^{-9}\times1}{10^{-3}}$

$= 0.5\ mm or 5 x 10^{-4}m$

$\beta_{0} = \frac{2\lambda\text{D}}{\text{a}} = 10 \beta$

$\text{a} = \frac{2\times500\times10^{-9}\times1}{10\times5\times10^{-4}}$

$? =2 \times 10^{−4}m\ or\ 0.2\ mm.$

STD 12 Science
Physics
Alternating Current
NCERT

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