CBSE Class 11-science Answered
An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and refractive index 1.5 is placed close to the mirror in the space between the object and the mirror. Find the position of the final image formed. Take the distance of the near surface of the slab from the mirror to be 1 cm
Asked by seeni2005 | 12 Jan, 2021, 21:30: PM
Expert Answer
Shift of Object distance due to glass slab = [ 1 - (1/μ) ] × d
where μ is refractive index of glass slab and d is thickness of glass slab
Shift of object distance = [ 1 - (2/3) ] × 3 = 1 cm
Hence object distance u as seen by mirror = ( 21 -1 ) = 20 cm
Image distance v from mirror is determined using mirror equation as given below
(1/v) + (1/u) = 1/f ........................ (1)
where f is focal length of mirror , f = half of radius of curvature = 20/2 = 10 cm
Image distance v is determined using eqn.(1)
( 1 / v ) + ( 1 / 20 ) = ( 1/ 10 )
( 1 / v ) = ( 1/10 ) - (1/20) = 1/20 or v = 20 cm
Since reflected ray is passing through slab to form image , we get a shift of +1 cm away from mirror
Hence image is formed at v' = 20 +1 = 21 cm
Answered by Thiyagarajan K | 13 Jan, 2021, 09:53: AM
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