Question

Geometrical Optics
UNIT-12
An object 10cm high is placed at a distance of 20cm from a concave lens of focal length 15 cm

Answer 1

Answer:

Given:

Focal length of concave mirror, f = -20 cm

Distance of object from the mirror, u = -10 cm

use formula, 1/v + 1/u = 1/f

or, 1/v = 1/f - 1/u

or, 1/v = 1/-20 - 1/-10

or, 1/v = -1/20 + 1/10 = 1/20

or, v = +20cm

hence, image position is the 20cm right side from the pole of the concave mirror as shown in the diagram. The image is straight and bigger than the object.

solution

Answer 2

Correct question

An object 10cm high is placed at a distance of 20cm from a concave lens of focal length 15 cm find the position of the image.

Given :

In concave lens,

Height of object = 10 cm

Object distance = 20 cm

Focal length = 15 cm

To find :

The position of the image

Solution :

using lens formula that is,

» The formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

$\boxed{ \bf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}}$

where,

• v denotes image distance
• u denotes object distance
• f denotes focal length

by substituting all the given values in the formula,

$\dashrightarrow \sf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}$

$\dashrightarrow \sf \dfrac{1}{v} - \dfrac{1}{ - 20}= \dfrac{1}{ - 15}$

$\dashrightarrow \sf \dfrac{1}{v} + \dfrac{1}{20}= \dfrac{1}{ - 15}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{1}{ - 15} - \dfrac{1}{20}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{ - 4 - 3}{60}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{ - 7}{60}$

$\dashrightarrow \sf \dfrac{1}{v} = - \dfrac{60}{7}$

$\dashrightarrow \sf v = 8.5 \: cm$

Thus, the position of image is 8.5 cm