Question

a) Draw a ray diagram to show the formation of image of an object placed between infinity and the optical centre of a Concave lens.
b) A concave lens of focal length 15 cm forms an image 10 cm from the lens. Calculate
i) the distance of the object from the lens
ii) the magnification for the image formed
iii) the nature of the image formed

Answer 1

the answer is mentioned in the pic

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Answer 2

Answer:

A concave lens always forms a virtual, erect image on the same side

of the object.

Image distance (v) = –10 cm;

Focal length (f) = –15 cm;

(i) Object distance (u) = ?

Since

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

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

$\frac{1}{u} = \frac{1}{ - 15} - ( \frac{1}{ - 10} )$

$\frac{1}{u} = \frac{ - 1}{30}$

u = – 30 cm

Thus, the object-distance is 30 cm.

(ii) Magnification m = v/u

$m = \frac{ - 10cm}{ - 30cm}$

$m = + \frac{1}{3}$

(iii) The image is erect and virtual and diminished.

Explanation:

For ray diagram

AB= Object

A'B'= Image