The correct option is
A (+30 cm, −14 mm)Reflection at concave mirror (M1):
Object distance,
u1=−20 cm
Focal distance,
f1=−15 cm
Using mirror formula,
1v1+1u1=1f1 We get,
1v1=1−15+120
⇒1v1=5−300
⇒v1=−60 cm
Magnification:
m1=−v1u1
⇒m1=−(−60)−20
⇒m1=−3
Since, initially the height of the point
P is
h=2 mm, so the height at which the image point will be
h′=|m1|×h
⇒h′=3×2
⇒h′=6 mm
Reflection at convex mirror (M2):
The image formed by concave mirror acts as an object for convex mirror. So,
object distance,
u2=(+60−50)=+10 cm
Focal length,
f2 = +20 cm
Using mirror formula:
1u2=1f2−1v2
⇒1u2=120−110
⇒1u2=−120
∴v2=−20 cm
Hence, the final position of image after two successive reflection at concave mirror and then at convex
20 cm in front of convex mirror,
M2.
Magnification:
m2=−v2u2=−(−20)10=2
So, the image of the point will be formed above
= m2 × (−6−2)
=2×(−8) mm
=−16 mm principal axis of
m2
So, the coordinate point of the final image will be
((50+v2) cm,(−16+2) mm)
=((50−20) cm,−14 mm)
=(30 cm,−14 mm)
Hence, option (a) is the correct answer.
Why this question:
To understand the image formation by combination of spherical mirrors. |