Question

23. Image of an object at infinity is formed by a convex lens of focal length
30 cm such that the size of the image is 12 cm. If a concave lens of focal
length 20 cm is placed in between the convex lens and the image, at a
distance 26 cm from the convex lens, size of the new image is
1) 2.5 cm
2) 2.0 cm
3) 1.025 cm 4) 1.05 cm​

Answer 1

Answer:

1)2.5cm

Explanation:

u=+4

u=+4 f=−20

u=+4 f=−20

1/v -1/u=1/f

1/v-1/4=-1/20

v=+5

v=+5size of image is

5/4 ×2=2.5

option1) is correct

Answer 2

CORRECT QUESTION:

Image of an object at infinity is formed by a convex lens of focal length 30 cm such that the size of the image is 2 cm. If a concave lens of focal length 20 cm is placed in between the convex lens and the image, at a distance 26 cm from the convex lens, size of the new image is.

ANSWER:

• Option 1) 2.5 cm

GIVEN:

• Convex lens of focal length 30 cm such that the size of the image is 2 cm.

• A concave lens of focal length 20 cm is placed in between the convex lens and the image, at a distance 26 cm from the convex lens.

TO FIND:

• Size of the new image.

EXPLANATION:

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

u = 30 - 26 = 4 cm

f = - 20 cm

$\sf \dfrac{1}{ - 20} = \dfrac{1}{v} - \dfrac{1}{4}$

$\sf \dfrac{1}{ - 20} +\dfrac{1}{4} = \dfrac{1}{v}$

$\sf \dfrac{1}{v} = \dfrac{ - 1 + 5}{ 20}$

$\sf \dfrac{1}{v} = \dfrac{4}{ 20}$

$\sf \dfrac{1}{v} = \dfrac{1}{5}$

$\sf v = 5 \ cm$

$\boxed{ \bold{ \large{ \gray{ \frac{v}{u} = \frac{h'}{h} }}}}$

v = 5 cm

u = 4 cm

h = 2 cm

$\sf\dfrac{5}{4} = \dfrac{h'}{2}$

$\sf\dfrac{5}{4} \times 2 = h'$

$\sf\dfrac{5}{2} = h'$

h' = 2.5 cm

HENCE THE HEIGHT OF THE NEW IMAGE WILL BE 2.5 cm.

QUESTION:

Image of an object at infinity is formed by a convex lens of focal length 30 cm such that the size of the image is 12 cm. If a concave lens of focal length 20 cm is placed in between the convex lens and the image, at a distance 26 cm from the convex lens, size of the new image is.

ANSWER:

• Size of the new image = 15 cm.

GIVEN:

• Convex lens of focal length 30 cm such that the size of the image is 12 cm.

• A concave lens of focal length 20 cm is placed in between the convex lens and the image, at a distance 26 cm from the convex lens.

TO FIND:

• Size of the new image.

EXPLANATION:

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

u = 30 - 26 = 4 cm

f = - 20 cm

$\sf \dfrac{1}{ - 20} = \dfrac{1}{v} - \dfrac{1}{4}$

$\sf \dfrac{1}{ - 20} +\dfrac{1}{4} = \dfrac{1}{v}$

$\sf \dfrac{1}{v} = \dfrac{ - 1 + 5}{ 20}$

$\sf \dfrac{1}{v} = \dfrac{4}{ 20}$

$\sf \dfrac{1}{v} = \dfrac{1}{5}$

$\sf v = 5 \ cm$

$\boxed{ \bold{ \large{ \gray{ \frac{v}{u} = \frac{h'}{h} }}}}$

v = 5 cm

u = 4 cm

h = 12 cm

$\sf\dfrac{5}{4} = \dfrac{h'}{12}$

$\sf\dfrac{5}{4} \times 12 = h'$

$\sf 5\times 3 = h'$

h' = 15 cm

HENCE THE HEIGHT OF THE NEW IMAGE WILL BE 15 cm.