Question

Subject: Physics
9334 ५
M.Marks :D
Q. When an object is placed at a distance of 60 cm from spherical diverging
mirror the magnification produced is 0.5. Where should be the object placed to
produce magnification of 1/3.
102 Marks)​

Answer 1

Answer:

u=−60cm

u

−v

=

2

1

⇒

−60

−v

=

2

1

⇒v=30cm

⇒

v

1

+

u

1

=

f

1

⇒

f

1

=

30

1

+

−60

1

⇒

f

1

=

60

2−1

=

60

1

⇒f=60cm

Now magnification=

3

1

u

−v

=

3

1

⇒v=

3

−u

∴

v

1

+

u

1

=

f

1

⇒

3

−u

1

+

u

1

=

f

1

⇒

u

−3

+

u

1

=

60

1

⇒

u

−2

=

60

1

⇒u=−120cm

∴ Object should be placed at 120cm.

Explanation:

u=−60cm

u

−v

=

2

1

⇒

−60

−v

=

2

1

⇒v=30cm

⇒

v

1

+

u

1

=

f

1

⇒

f

1

=

30

1

+

−60

1

⇒

f

1

=

60

2−1

=

60

1

⇒f=60cm

Now magnification=

3

1

u

−v

=

3

1

⇒v=

3

−u

∴

v

1

+

u

1

=

f

1

⇒

3

−u

1

+

u

1

=

f

1

⇒

u

−3

+

u

1

=

60

1

⇒

u

−2

=

60

1

⇒u=−120cm

∴ Object should be placed at 120cm.