Question

An object is places 12.5 cm in front of a diverging lens with a focal length of 10 cm.
Determine the image distance.​

Answer 1

Explanation:

$12.5 \times 10$

Answer 2

Answer:

Given, object distance, u = –25 cm

Focal length, f = –25 cm, image distance, v = ?

From lens formula,

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

v

1

−

u

1

=

f

1

So,

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

v

1

=

f

1

+

u

1

\frac{1}{u} = \frac{u + f}{uf}

u

1

=

uf

u+f

v = \frac{uf}{u + f}v=

u+f

uf

v = \frac{ - 25 \times ( - 25)}{ - 25 - 25} = - 12.5 \: cmv=

−25−25

−25×(−25)

=−12.5cm

Nature of image

1. Virtual.

2. Erect.

3. On the same side of lens.

4. Diminished.

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