Question

the single slit diffraction arrangement is completely immersed in water.the fringe system ? ​

Answer 1

Answer:

A single slit diffraction pattern is completely immersed in water without changing any other parameter. How is the width of central maximum affected? Answer: Solution : Wave length of light in water (λ′=λ/μ) decreases

Answer 2

Answer:

The single slit diffraction arrangement is completely immersed in water then the central fringe width, fringe width, and angular fringe width of the system decrease by 1/μ times.

Explanation:

In the single-slit diffraction experiment, we can observe the bending of light or diffraction that causes light from a coherent source to interfere and produce a distinctive pattern on the screen called the diffraction pattern. Diffraction is evident when the sources are small enough that they are relatively the size of the wavelength of the light.

Fringe width of central maxima,

$(W)=\frac{2\lambda D}{d}$     (1)

Where,

λ=wavelength of light

D= distance between the screen and the slit

d=distance between the slit

And,

Width of fringes other than central,

$\beta=\frac{\lambda D}{d}$        (2)

Angular fringe width,

$\theta$$=\frac{2\lambda}{d}$      (3)

Now since the diffraction arrangement is wholly immersed in water its wavelength will change i.e.

$\lambda'=\frac{\lambda}{\mu}$    (4)

λ'= wavelength of light in the new medium

μ=refractive index of water

by substituting equation (4) in equations (1),(2), and (3) we get;

$W'=\frac{2\lambda D}{\mu d}$$=\frac{W}{\mu}$    (5)

$\beta'=\frac{\lambda D}{\mu d}$$=\frac{\beta}{\mu}$         (6)

$\theta'=\frac{2\lambda}{\mu d}$$=\frac{\theta}{\mu}$      (7)

Hence, central fringe width, fringe width other than central angular fringe width decreases by 1/μ times when immersed completely in water.

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