Question

in a single slit diffraction experiment light of wavelength 6000 A is used and the screen is 50 cm away from the slit . if distance between the first and third minima is 3 mm, width of the slit is​

Answer 1

Explanation:

now just putbthe values and solve it

Answer 2

The slit width is $2\times 10^{-4}\ m$.

Explanation:

Given that,

Wavelength of light used in single slit diffraction experiment, $\lambda=6000\ A=6\times 10^{-7}\ m$

Distance from the screen to the slit, D = 50 cm = 0.5 m

The distance between the first and third minima is 3 mm

We need to find the width of the slit. Th position of minima in the diffraction pattern is given by :

$x_n=\dfrac{n\lambda D}{d}$

Therefore,

$x_3-x_1=(3-1) \dfrac{D\lambda}{d}\\\\x_3-x_1=2 \dfrac{D\lambda}{d}\\\\d=2 \dfrac{D\lambda}{x_3-x_1}\\\\d=2\times \dfrac{0.5\times 6\times 10^{-7}}{3\times 10^{-3}}\\\\d=2\times 10^{-4}\ m$

So, the slit width is $2\times 10^{-4}\ m$.

Learn more,

Single slit diffraction

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