Question

In a Young's double slit experiment, if there is no
initial phase difference between the light from the
two slits, a point on the screen corresponding to
the fifth minimum has path difference​

Answer 1

Answer:

Path difference of light for fifth minimum on the screen is given as

$\Delta x = \frac{9}{2} \lambda$

Explanation:

As we know that for the minimum intensity of light on the screen we will phase difference given as

$\phi = (2N + 1)\pi$

now the relation between phase difference and path difference is given as

$\phi = \frac{2\pi}{\lambda} \Delta x$

so we have

$(2N + 1) \pi = \frac{2\pi}{\lambda} \Delta x$

now for fifth minimum intensity we can say

$N = 4$

so we have

$9\pi = \frac{2\pi}{\lambda} \Delta x$

$\Delta x = \frac{9}{2} \lambda$

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Topic : Young's Double slit experiment

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Answer 2

Answer:

for to calculate path diffrence in minimum

∆x= (2n-1) × lambda/2

putting value of n you get answer