Question

a concave lens of focal length 60cm is used to form an image of an object of length 10 centimetre kept at a distance of 30 cm from it used lens formula to determine the nature position and length of the image formed also draw labelled ray diagram to show the image formation in the case above

Answer 1

Answer:

hope this helps u

Explanation:

the above one answer is 20cm

Answer 2

$\bold{Answer\;:-}$

Given :-

› Focal length of the concave lens, f = - 60 cm

› Object distance, u = - 30 cm

› Object size, h = 10 cm

Now, by using the formula,

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

• $\frac{1}{-60}$ = $\frac{1}{v}$ + $\frac{1}{-30}$

• $\frac{1}{-60}$ + $\frac{1}{-30}$ = $\frac{1}{v}$

• $\frac{1}{v}$ = $\frac{3}{-60}$

• $\frac{1}{v}$ = $\frac{1}{-20}$

• v = 20 cm

Now, by using the formula,

• m = $\frac{v}{u}$ = $\frac{h'}{h}$

• $\frac{-20}{-30}$ = $\frac{h'}{10}$

• $\frac{2}{3}$ = $\frac{h'}{10}$

• h' = $\frac{2}{3}$ × 10

• h' = 6.6 cm

Hence, nature of the image is virtual, erect and diminished. The image is between the F1 and the optical centre of the lens and the length of the image is 6.6 cm.

$\bold{Hope\;it \; helps\;!}$