Question

If the eighth bright band due to light of wavelengin
coincides with ninth bright band from light of wavelength
de in Young's double slit experiment, then the possible
wavelengths of visible light are J & K CET 2010)
(a) 400 nm and 450 nm (b) 425 nm and 400 nm
(c) 400 nm and 425 nm (d).950 nm and 400 nm
VOT​

Answer 1

Answer:

Wavelength of visible light that will show the same is 400 nm and 450 nm

Explanation:

As we know that 8th bright fringe coincide with 9th bright fringe of other wavelength

Now we know that the condition of maximum intensity on screen is given as

$\Delta x = N\lambda$

now we have

$N_1\lambda_1 = N_2\lambda_2$

so we have

$8\lambda_1 = 9\lambda_2$

$\lambda_1 : \lambda_2 = 9 : 8$

so it must be 400 nm and 450 nm

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Topic : Interference of light

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