Question

A pin is placed 10cm in front of a convex lens of focal length 20cm made of material of refractive index

1.5 . The surface of the lens farther away from the pin is silvered and has a radius of curvature 22cm . Determine the

position of the final image. Is the image real or virtual?

Answer 1

Answer:

The image will be formed 11 cm in front of concave mirror and its real.

Explanation:

Given that,

Focal length f = 20 cm

Object distance u = -10 cm

Refractive index n = 1.5

Radius of curvature of silvered lens R = 22 cm

The surface of the lens is silvered then lens will behave as a concave mirror.

So, the focal length of concave mirror

$f_{m} = -\dfrac{R}{2}$

$f_{m} = - 11 cm$

Therefore, the power of the mirror

$P_{m} = -\dfrac{1}{f_{m}}$

$P_{m} = -\dfrac{1}{11} D$

Further, focal length of the lens is 20 cm.

So, the power of lens is

$P = \dfrac{1}{20} D$

When the light after passing through the lens will be reflected back by concave mirror through convex lens again then, image will formed

Now, the power is

$P = P_{l} + P_{m} + P_{l}$

$P = 2\times \dfrac{1}{20} + \dfrac{1}{11}$

$P = \dfrac{21}{110} D$

The focal length of equivalent mirror

$f = -\dfrac{110}{21} cm$

Now, the object placed in front of a convex lens

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$-\dfrac{21}{110} = \dfrac{1}{v} - \dfrac{1}{10}$

$v = -11 cm$

Hence, the image will be formed 11 cm in front of concave mirror and its real.

Answer 2

Explanation:

Given that,

Focal length f = 20 cm

Object distance u = -10 cm

Refractive index n = 1.5

Radius of curvature of silvered lens R = 22 cm

The surface of the lens is silvered then lens will behave as a concave mirror.

So, the focal length of concave mirror

f_{m} = -\dfrac{R}{2}fm=−2R

f_{m} = - 11 cmfm=−11cm

Therefore, the power of the mirror

P_{m} = -\dfrac{1}{f_{m}}Pm=−fm1

P_{m} = -\dfrac{1}{11} DPm=−111D

Further, focal length of the lens is 20 cm.

So, the power of lens is

P = \dfrac{1}{20} DP=201D

When the light after passing through the lens will be reflected back by concave mirror through convex lens again then, image will formed

Now, the power is

P = P_{l} + P_{m} + P_{l}P=Pl+Pm+Pl

P = 2\times \dfrac{1}{20} + \dfrac{1}{11}P=2×201+111

P = \dfrac{21}{110} DP=11021D

The focal length of equivalent mirror

f = -\dfrac{110}{21} cmf=−21110cm

Now, the object placed in front of a convex lens

\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}f1=v1+u1

-\dfrac{21}{110} = \dfrac{1}{v} - \dfrac{1}{10}−11021=v1−101

v = -11 cmv=−11cm

Hence, the image will be formed 11 cm in front of concave mirror and its real.