Question

a concave lens of focal length 15 cm forms an image 10 cm from the lens.how far is the object placed from the lens

Answer 1

This question is solved by Len's Formula

Answer 2

Answer:

Focal length (f ) = -15 cm

Focal length (f ) = -15 cmDistance of image (v) = -10 cm

Focal length (f ) = -15 cmDistance of image (v) = -10 cm\(\frac{1}{v} – \frac{1}{u} = \frac{1}{f}\)

Focal length (f ) = -15 cmDistance of image (v) = -10 cm\(\frac{1}{v} – \frac{1}{u} = \frac{1}{f}\)1/u = -(1/10) – (1/- 15)

Focal length (f ) = -15 cmDistance of image (v) = -10 cm\(\frac{1}{v} – \frac{1}{u} = \frac{1}{f}\)1/u = -(1/10) – (1/- 15)1/u =1/15 – 1/10

Focal length (f ) = -15 cmDistance of image (v) = -10 cm\(\frac{1}{v} – \frac{1}{u} = \frac{1}{f}\)1/u = -(1/10) – (1/- 15)1/u =1/15 – 1/101/u = -0.033

Focal length (f ) = -15 cmDistance of image (v) = -10 cm\(\frac{1}{v} – \frac{1}{u} = \frac{1}{f}\)1/u = -(1/10) – (1/- 15)1/u =1/15 – 1/101/u = -0.033u = -30 cm

Focal length (f ) = -15 cmDistance of image (v) = -10 cm\(\frac{1}{v} – \frac{1}{u} = \frac{1}{f}\)1/u = -(1/10) – (1/- 15)1/u =1/15 – 1/101/u = -0.033u = -30 cmSo, the object is placed 30 cm away from the concave lens.