Wave OpticsHard

Question

A fish in an aquarium, 30 cm deep in water can see a light bulb kept 50 cm above thesurface of water. The fish can also see the image of this bulb in the reflecting bottom surface of the aquarium.Total depth of water is 60 cm. then the appartent distance be tween the two images the apparent distance between the two images seen by the fish is (μw = 4/3)

Options

A.140 cm
B. cm
C. cm
D. cm

Solution

       
Appartent distance of the bulb of from the fish d1 = 50μ + 30
apparemt distance of the image
d2 = 50μ + 60 + 30
∴ d1 + d2 = 100μ + 120
    = + 120 = cm
    = 253.3cm

Create a free account to view solution

View Solution Free
Topic: Wave Optics·Practice all Wave Optics questions

More Wave Optics Questions

Two waves of intensties I and 4I superimposes :...In Young′s double slit experiment, the separation between the slits is halved and the distance the slits and the s...Bending of light waves at the sharp edges of an opaque obstacle is known as...If radius of first H.P.Z. is 0.5 mm. on zone plate and wavelength of light 500 nm is used. Then distance of maximum brig...The resultant amplitude in interference with two coherent sources depends upon :...