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A light bulb is at a depth of D below th...

A light bulb is at a depth of D below the surface of water. An opaque disc of radius R is placed on the surface of water just above the bulb. The bulb is not at all seen through the surface of water, then ( n = Refractive index of water)

A

`R = (D)/(sqrt(n^(2) -1))`

B

`R gt (D)/(sqrt(n^(2) - 1))`

C

`R lt (D)/(sqrt(n^(2) - 1))`

D

`R = D sqrt(n^(2) - 1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`sin i_(c ) = (1)/(n)`
`rArr " " (R )/(sqrt(R^(2) + D^(2) ) ) = (1)/(n)`
`rArr " " (R^(2) + D^(2))/(R^(2)) = n^(2)`
`rArr " " 1 + (D^(2))/(R^(2)) = n^(2) rArr (D^(2))/(R^(2)) = n^(2) = 1 `
`rArr " " R = (D)/(sqrt(n^(2) -1))`
If bulb is not seen, `R gt (D)/(sqrt(n^(2) -1))`
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