Question

Fringe width for an interference pttern become k times its value if the incident wavelenght changes from 10 to 20 then value of k is

Answer 1

Answer:

Explanation:

The two sources should emit, continuously, waves of some wave-length or frequency. ... The fringe width varies inversely as distance 'd' between the two sources. So, interference pattern will be more clear and distant if 'd' is small. ... Let d be the distance between two coherent sources A and B of wavelength

Answer 2

Value of k is 2.

Explanation:

We know,

Fringe width of an interference pattern is is given by , $\beta=\dfrac{\lambda D}{d}$

Initially when , $\lambda=10$

$\beta=\dfrac{10D}{d}$     ......equation 1.

Now, it is given that when , $\lambda=20$ fringe width becomes K times the initial fringe width.

Therefore, $k\times \beta=\dfrac{20D}{d}$     ......equation 2.

$k\times \beta=2\times\dfrac {10 D}{d}$

putting value of $\dfrac{10D}{d}=\beta$  in above equation.

We get,

k = 2 .

Therefore, value of k is 2.

Learn more :

Interference.

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