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ICSE Class 9 Answered

An object is placed 42cm from a concave mirror of curvature 16 cm. A glass slab of thickness 6cm and R.I. 3/2 is then placed between the object and mirror. Find the position of the final image formed (the near surface of the slab from the mirror is 1cm)
Asked by u.sudhishna | 05 Aug, 2022, 15:51: PM
answered-by-expert Expert Answer
Object is 42 cm in front of concave mirror .
There is a glass slab of thickness 6 cm in front of mirror such that near surface of glass slab is 1 cm from concave mirror.
 
Hence light ray has to travel a distance 36 cm in air and 6 cm in glass slab .
 
If refractive index of glass slab is n , then equivalent air distance of light path in glass slab = n × t ,
where t is thickness of glass slab
 
hence for 6 cm thick glass slab, air distance = 6 × 1.5 = 9 cm
 
Hence total air distance travelled by light before getting reflected by concave mirror is ( 36 + 9 ) cm = 45 cm
 
we have mirror equation as
 
(1/v ) + ( 1/u ) = 1/f
 
where v is mirror-to-image distabnce , u is mirror-to-object distance and f is focal length
 
since radius of curavture of mirror is 16 cm , focal length f = 8 cm
 
By using cartesian sign convention, we write mirror equation as
 
(1/v) - (1/45) = -1/8
 
(1/v) = (1/45) - ( 1/8 )
 
from above expression , we get v = - 9.7 cm , i.e 9.7 cm in front of mirror
 
since glass slab surface is 1 cm in front of mirror , then equivalent air distance 8.7 cm will be travelled in glass slab
 
distance travelled in glass slab = 8.7 / 1.5 = 5.8 cm
 
Hence image is formed infront of mirror at a distance ( 1 cm in air + 5.8 cm in glass slab )
hence image is formed inside glass slab at a distance 6.8 cm from mirror
Answered by Thiyagarajan K | 06 Aug, 2022, 00:05: AM

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