Question

25. The wavelength of light in vacuum is 5000A
When it travels normally through diamond of
hickness 1.0mm find the number of waves of light
in 1.0mm of diamond.
(Refractive index of diamond = 2,417)
1) 4834 waves
2) 5834 waves
3) 4384 waves
4) 6834 waves
​

Answer 1

Answer:

No.of waves travelled in diamond = distance travelled in diamond / wavelength

No.of waves = μd / v

= 2.417 * 1 * 10⁻³ / 5000 * 10⁻¹⁰

= 4834 waves

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