Question

The image of candle flame placed at a distance of 40cm from a spherical lens is formed on a screen placed on the other side of the lens at a distance of 40 cm from the lens.
(a) Identify the type of lens and compute its focal length.
(b) Write the nature of the image formed if the candle flame is shifted 25cm towards the
lens.

Answer 1

(a) The type of the lens is convex lens and its focal length is 20 cm.

(b) The nature of the image is virtual, erect and enlarged.

Given:

Object distance = u = - 40 cm

Image distance = v = 40 cm

To find:

Type of the lens = ?

Focal length = ?

Nature of the image after the shift = ?

Formula used:

Lens formula:

$\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}}$

Magnification formula:

$\bold{m=\frac{v}{u}}$

Solution:

(a) Focal length is:

$\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}}$

$\bold{\frac{1}{f}=\frac{1}{40}-\frac{1}{(-40)}}$

$\bold{\frac{1}{f}=\frac{1}{40}+\frac{1}{40}}$

$\bold{\frac{1}{f}=\frac{2}{40}=\frac{1}{20}}$

$\bold{\therefore f= 20 \ cm}$

Since, the focal length has positive sign, it it a convex lens.

(b) Nature of the image formed when candle flame is shifted 25 cm:

Distance of the object = $\bold{-(40-25)=-15 \ cm}$

Now, image distance is:

$\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}}$

$\bold{\frac{1}{20}=\frac{1}{v}-\frac{1}{(-15)}}$

$\bold{\frac{1}{20}=\frac{1}{v}+\frac{1}{(15)}}$

$\bold{\frac{1}{20}-\frac{1}{15}=\frac{1}{v}}$

$\bold{\frac{1}{v}=\frac{3-4}{60}=-\frac{1}{60}}$

$\bold{\therefore v=-60 \ cm}$

Now, we need to find the magnification:

$\bold{m=\frac{v}{u}}$

$\bold{m=\frac{-60}{-15}=4}$

Since, the magnification is positive, then the image is virtual, erect and enlarged.

Answer 2

Explanation:

(a) The type of the lens is convex lens and its focal length is 20 cm.

(b) The nature of the image is virtual, erect and enlarged.

Given:

Object distance = u = - 40 cm

Image distance = v = 40 cm

To find:

Type of the lens = ?

Focal length = ?

Nature of the image after the shift = ?

Formula used:

Lens formula:

\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}}

f

1

=

v

1

−

u

1

Magnification formula:

\bold{m=\frac{v}{u}}m=

u

v

Solution:

(a) Focal length is:

\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}}

f

1

=

v

1

−

u

1

\bold{\frac{1}{f}=\frac{1}{40}-\frac{1}{(-40)}}

f

1

=

40

1

−

(−40)

1

\bold{\frac{1}{f}=\frac{1}{40}+\frac{1}{40}}

f

1

=

40

1

+

40

1

\bold{\frac{1}{f}=\frac{2}{40}=\frac{1}{20}}

f

1

=

40

2

=

20

1

\bold{\therefore f= 20 \ cm}∴f=20 cm

Since, the focal length has positive sign, it it a convex lens.

(b) Nature of the image formed when candle flame is shifted 25 cm:

Distance of the object = \bold{-(40-25)=-15 \ cm}−(40−25)=−15 cm

Now, image distance is:

\bold{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}}

f

1

=

v

1

−

u

1

\bold{\frac{1}{20}=\frac{1}{v}-\frac{1}{(-15)}}

20

1

=

v

1

−

(−15)

1

\bold{\frac{1}{20}=\frac{1}{v}+\frac{1}{(15)}}

20

1

=

v

1

+

(15)

1

\bold{\frac{1}{20}-\frac{1}{15}=\frac{1}{v}}

20

1

−

15

1

=

v

1

\bold{\frac{1}{v}=\frac{3-4}{60}=-\frac{1}{60}}

v

1

=

60

3−4

=−

60

1

\bold{\therefore v=-60 \ cm}∴v=−60 cm

Now, we need to find the magnification:

\bold{m=\frac{v}{u}}m=

u

v

\bold{m=\frac{-60}{-15}=4}m=

−15

−60

=4

Since, the magnification is positive, then the image is virtual, erect and enlarged.