Question

3 points
Q 11. An object is kept at a
distance of 60 cm from a
convex lens of focal length 20
cm. Find the position and
nature of image formed. If the
object has a size of 4 cm, find
the size of image formed. *
o
4 cm
O zem
от cm​

Answer 1

U = 60

f= 20

1/f= 1/v-1/u

1/20 = 1/v - 1/60

1/20= 1/v -1/60

1/20+1/60=1/v

3+1/60 = 1/v

4/60=1/v

60/4 =v

15=v

nature of the image virtual and errect

m = h'/h= V/u

m= h'/4= 15/60

m = h'= 15×4/60

m = h' = 1

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Answer 2

GIVEN:

• Object distance, u = - 60 cm

• Focal length of convex lens , f = + 20 cm

• Size of object, h = 4 cm

TO FIND:

• Position of image ( image distance) ,v
• Nature of image.
• Size of image, H

FORMULAE:

• According to lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

• Magnification of lens, m = v/u = H/h

SOLUTION:

STEP 1: To find image distance.

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ \frac{1}{20} = \frac{1}{v} - \frac{1}{ (- 60)} \\ \\ \frac{1}{20} = \frac{1}{v} + \frac{1}{60} \\ \\ \frac{1}{20} - \frac{1}{60} = \frac{1}{v} \\ \\ \frac{1}{v} = \frac{60 - 20}{20 \times 60} \\ \\ \frac{1}{v} = \frac{40}{20 \times 60} \\ \\ v = \frac{20 \times 60}{40} \\ \\ v = \frac{60}{2} \\ \\ v = 30 \: cm$

Image distance is 30 cm.

Position of image is 30 cm to the right of convex lens.

Since the image is formed at the other side to the object, the rays actually meet and forms real image.

STEP 2: To find size of image

$\frac{v}{u} = \frac{H}{h} \\ \\ \frac{30}{ - 60} = \frac{H}{4} \\ \\ - \frac{1}{2} = \frac{H}{4} \\ \\ H = - \frac{4}{2} \\ \\ H = - 2 \: cm$

Size of image is 2 cm.

Negative sign of size of image indicates that the image formes is inverted.

ANSWER:

• Position of image is 30 cm to the right of convex lens.

• Size of image is 2 cm.

• It is a real inverted image.