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.
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Answer:
The shift in position of object due to slab is given by
Δt=t(1−
μ
1
)=3(1−
2
3
1
)=3(1−
3
2
)=1cm
The apparent distance of object from mirror 21−1=20cm
The position of image formed by concave mirror can be found by mirror formula
v
1
+
u
1
=
f
1
...(i)
As object and focus of mirror lies left side of pole of mirror
u=−20,f=−
2
R
=−5
substituting values in equation (i)
1
+
−20
1
=
−5
1
v
1
=
20
1
−
20
4
v=−
3
20
As image is other side of the slab,the image will be again shifted by 1 cm away from mirror.
Final distance of image from mirror=
3
20
+1cm=7.67cm
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