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प्रश्न
| A ray of light travels from a denser to a rarer medium. After refraction, it bends away from the normal. When we keep increasing the angle of incidence, the angle of refraction also increases till the refracted ray grazes along the interface of two media. The angle of incidence for which it happens is called critical angle. If the angle of incidence is increased further the ray will not emerge and it will be reflected back in the denser medium. This phenomenon is called total internal reflection of light. |
A point source of light is placed at the bottom of a tank filled with water, of refractive index µ, to a depth d. The area of the surface of water through which light from the source can emerge is:
विकल्प
`(π"d"^2)/(2(µ^2 - 1))`
`(π"d"^2)/((µ^2 - 1))`
`(π"d"^2)/(sqrt2 sqrt(µ^2 - 1))`
`(2π"d"^2)/((µ^2 - 1))`
उत्तर
`bb(("πd"^2)/((µ^2 - 1)))`
Explanation:
r = d tan θc
sin θc = `1/µ`
∴ cos θc = `sqrt(1 - 1/µ^2)`
∴ r = `"d"(1//µ)/(sqrt(µ^2 - 1)/µ) = "d"/sqrt(µ^2 - 1)`
∴ area = πr2 = `(π"d"^2)/(µ^2 - 1)`
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| A ray of light travels from a denser to a rarer medium. After refraction, it bends away from the normal. When we keep increasing the angle of incidence, the angle of refraction also increases till the refracted ray grazes along the interface of two media. The angle of incidence for which it happens is called critical angle. If the angle of incidence is increased further the ray will not emerge and it will be reflected back in the denser medium. This phenomenon is called total internal reflection of light. |
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