Question

The projection lens of a projector has focal length 5 cm. It is desired to get an image with a magnification 30. The distance of the screen from the lens must be?
a)0.3 m

b)0.8 m

c)1.55 m

d)2.55

Answer 1

Given:

Projection lens has focal length 5 cm. Magnification required is 30.

To find:

Distance of screen from lens ?

Calculation:

Ae know that : magnification of lens is equal to ratio of image distance and object distance.

• Also for real images, magnification is always negative (i.e. -30).

$\rm \: magnification = \dfrac{v}{u}$

$\rm \implies - 30 = \dfrac{v}{ - 5}$

$\rm \implies v = 30 \times 5$

$\rm \implies v = 150 \: cm$

$\rm \implies v = 1.5 \: m$

$\rm \implies v \approx 1.55 \: m$

So, distance of screen from lens is 1.55 cm.

Answer 2

Answer:

c) 1.55m

Explanation:

Given

f = 5cm

m = -30 (∵ the image formed is real )

Calculation

acc. to lens formula

1/f = 1/v - 1/u

1/5 = 1/v - 1/u

1/u = 1/v - 1/5 ... (1)

m = v/u

-30 = v × 1/v - 1/5

-30 = v ( 5 - v / 5v)

-30 = 5v - v² / 5v

-30 = v ( 5 - v ) / v5

-30 × 5 = 5 - v

- 155 = -v

v = 155 ⇒ v = 1.55 m