Physics, asked by nasreenmemon2, 1 year ago

A concave mirror of radius 40 cm lies on the horizontal table and water is filled in it joto height of 5 cm a small dust particle floats on water surface vertically above the point of contact of the mirror with the table location of the image of the dust as seen from a point directly above it

Answers

Answered by poonambhatt213
4

Answer:

Explanation:

=> For the image formed by concave mirror:

object distance, u = -5 cm.

radius, R = -40 (suppose the image formed by concave mirror is vertically upward)

=> By using mirror equation,

1/u + 1/v = 2/R

1/v = 2/R - 1/u

1/v = 2/-40 - 1/-5

1/v = 6/40 cm

∴ v = 6.67 cm

=> This image I1 is formed at 6.67 cm behind the mirror. It is virtual image.

=> here, the reflected rays are reflected by the water surface and reach the viewer.

the deepness of the point I1 from the surface is,

6.67 cm + 5 cm = 11.67 cm

=> the image I1 will be shifted above due to the refraction at water surface,

distance of image formed due to refraction: 11.67 * 1-1/1.33 = 2.92 cm.

=> therefore, the final image is formed at 11.67 cm - 2.92cm = 8.75 cm beneath the water surface.

here, 8 + a/4 = 8.75

a/4 = 8.75 - 8

a/4 = 0.75

a = 0.75 * 4

∴ a = 3

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