Question

an object is placed from 50cm from the pole of a concave mirror of radius of curvature 40 cm. if the size of the object is 6 cm then find the nature position and size of the image​

Answer 1

12th

Physics

Ray Optics and Optical Instruments

Combination of Lenses

An object is placed exactly...

PHYSICS

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Asked on December 27, 2019 by

Uttam Rawat

An object is placed exactly midway between a concave mirror of radius of curvature 40cm and a convex mirror of radius of curvature 30cm. The mirrors face each other and are 50cm apart. Determine the nature and position of the image formed by successive reflections first at the concave mirror and then at the convex mirror.

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ANSWER

Given: An object is placed exactly midway between a concave mirror of radius of curvature 40cm and a convex mirror of radius of curvature 30cm. The mirrors face each other and are 50cm apart.

To find the nature and position of the image formed by successive reflections first at the concave mirror and then at the convex mirror.

Solution:

For concave mirror,

As per the given condition,

object distance, u=25cm, as the mirrors are 50cm apart.

Radius of curvature, R=40cm

So the focal length, f=

2

R

=

2

40

=20cm

Applying lens formula, we get

f

1

=

v

1

+

u

1

⟹

v

1

=

f

1

−

u

1

⟹

v

1

=

20

1

−

25

1

⟹

v

1

=

100

5−4

⟹v=100cm

Therefore the image is 100cm behind the concave mirror.

Magnification, m

1

=−

u

v

=−

25

100

=−4

Hence the image formed by the concave mirror is inverted, virtual and enlarged.

This becomes object for convex mirror, so

object distance, u

′

=−(100−50)=−50cm

Radius of curvature of convex mirror, r=30cm

So the focal length, f

′

=

2

r

=

2

30

=−15cm

Applying lens formula, we get

f

′

1

=

v

′

1

+

u

′

1

⟹

v

′

1

=

f

′

1

−

u

′

1

⟹

v

′

1

=

−15

1

−

−50

1

⟹

v

′

1

=

150

−10+3

⟹v=−21.4cm

Hence the position of the final image is 21.4 behind the convex mirror.

Magnification, m

2

=−

u

′

v

′

=−

−50

−21.4

=−0.43

So the nature of the final image formed is inverted of inverted, i.e., upright.

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Answer 2

Answer:

object distance=p=50cm

Radius of curvature=R=60cm

then f=R/2

f=60/2

f=30cm

image distance=q=?

using mirror formula

1/f=1/p+1/q

1/30=1/50+1/q

1/30–1/50=1/q

(5–3)/150=1/q

2/150=1/q

2q=150

q=75

image distance=75cm

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