JEE Class main Answered
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Asked by sarveshvibrantacademy | 10 May, 2019, 10:37: AM
Expert Answer
If t is thickness of glass slab of refractive index μ, it produces a shift t[ 1 -(1/μ) ] in the path of travelling direction.
Hence slab of thickness 3 cm with refractive index (3/2) produes a shift 3[1-(2/3)] = 1 cm and
the object-to-mirror distance 16 cm becomes 15 cm because of glass slab.
virtual image is formed at a distance v behind the mirror as per mirror equation (1/u)+(1/v) = (1/f)
Hence (-1/15)+(1/v) = (1/10) or v = 6 cm behind the mirror.
now object-to-image distance 16+6 = 22 cm is getting shifted due glass slab by 1 cm, hence the distance becomes 21 cm
if d cm is the distance between front surface of glass slab and object, we get image
due to partial reflection d cm behind the front surface of glass slab.
If two images are getting overlapped, then we have 2d = 21 or d = 10.5 cm
Answered by Thiyagarajan K | 11 May, 2019, 11:21: AM
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