> Q. In Young’s double slit experiment, the $10^{th}$ maximum of wavelength $lambda _1$ is at a distance of $y_1$ from the central maximum. When the wavelength of the source is changed to $lambda _2, 5^{th}$ maximum is at a distance of $y_2$ from its central maximum. The ratio $left(frac{y_{1}}{y_{2}}right)$ is – LIVE ANSWER TODAY

Q. In Young’s double slit experiment, the $10^{th}$ maximum of wavelength $lambda _1$ is at a distance of $y_1$ from the central maximum. When the wavelength of the source is changed to $lambda _2, 5^{th}$ maximum is at a distance of $y_2$ from its central maximum. The ratio $left(frac{y_{1}}{y_{2}}right)$ is

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In Young’s double slit experiment, the $10^{th}$ maximum of wavelength $lambda _1$ is at a distance of $y_1$ from the central maximum. When the wavelength of the source is changed to $lambda _2, 5^{th}$ maximum is at a distance of $y_2$ from its central maximum. The ratio $left(frac{y_{1}}{y_{2}}right)$ is

Solution:

The distance of $10^{th}$ maximum of wavelength $lambda _1$ from the central maximum is

$y_{1}=10lambda_{1} frac{D}{d}$

where $D$ is the distance of the slits from the screen and $d$ is the distance between the slits.

The distance of $5^{th}$ maximum of wavelength $lambda_{2}$ from the central maximum is

$y_{2}=5lambda_{2} frac{D}{d}$

$therefore frac{y_{1}}{y_{2}}=frac{2lambda_{1}}{lambda_{2}}$

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