Question

An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm.
(i) What is the nature of image?
(ii) What is the position of image?

Answer 1

Answer:The nature of image is dimenished and real and inverted the image is formed between focus2 and 2F2

Explanation:

When the object is placed behind 2F1 the paralell line from its tip get refracted in concave lens and passes through focus2 and the ray passes through optic centre it reterns so image is formed there

Answer 2

Image is Virtual & Erect

Explanation:

(a) Given

• Object distance (u) = -100 cm

• Focal length (f) = 50 cm

• Image distance = v

By lens formula;

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

$\frac {1}{v} - \frac {1}{-100} = \frac {1}{50}$

$\frac {1}{v} + \frac {1}{100} = \frac {1}{f}$

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

$\frac {1}{v} = \frac {2-1}{100} = \frac {1}{100}$

v = 100 cm

Hence the image is formed at 100 cm behind the lens.

• Object distance (u) = -100cm

• Focal length (f) = -25cm

• Image distance = v

By lens formula;

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

$\frac {1}{v} - \frac {1}{-100} = \frac {-1}{25}$

$\frac {1}{v} = \frac {-1}{25} - \frac {1}{100}$

$\frac {1}{v} = \frac {-4-1}{100} = \frac {-5}{100}$

v = $\frac {-100}{5}$ = - 20 cm

• b) For 1st case

• Object distance (u) = -100cm

• Image distance (v) = 100cm.

• By Magnification Formula

Magnification (m) = $\frac {v}{u}$

Magnification, m = $\frac {100}{-100}$

Therefore, Magnification for the 1st case is -1. The negative sign means the image is real and inverted.

• For 2nd case

• Object distance (u) = -100cm

• Image distance (v) = -20cm.

• By Magnification Formula

Magnification (m) = $\frac {v}{u}$

Magnification, m = $\frac {-20}{-100} = \frac {1}{5}$

= 0.2

Therefore, Magnification for 2nd case is 0.2. The positive sign means the image is erect and virtual.