Question

A concave lens has a focal length 15cm . At what distance should the object be placed from the lens so that it forms an image at 10cm from the lens ? What will be the nature and magnification of the image ?​

Answer 1

Answer:

• Object should be placed at a distance of 30 cm in front of the lens

• magnification of image = + 1/3

• Nature of Image : virtual , erect and diminished

Explanation:

focal length f = - 15 cm (∵ concave lens)

Image distance v = - 10 cm ( ∵  image is on same side as that of object)

From Lens formula :

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

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

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

  = - $\frac{1}{30}$

u = - 30 cm

∴ object should be placed at a distance of 30 cm (At 2F₁) in front of the lens.

magnification m = v/u = -10/(-30) = + 1/3 = + 0.333

Since m is positive : Image is virtual and erect

Since |m| < 1 : Image is diminished

Answer 2

Answer:-

Object should be placed at a distance of 30 cm in front of the lens

magnification of image = + 1/3

Nature of Image : virtual , erect and diminished

Explanation:

focal length f = - 15 cm (∵ concave lens)

Image distance v = - 10 cm ( ∵ image is on same side as that of object)

From Lens formula :

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} </p><p>\frac{-1}{10} - \frac{1}{u} = - \frac{1}{15}</p><p>\frac{1}{u} = \frac{1}{15} - \frac{1}{10} </p><p> = - \frac{1}{30}$

u = - 30 cm

∴ object should be placed at a distance of 30 cm (At 2F₁) in front of the lens.

magnification m = v/u = -10/(-30) = + 1/3 = + 0.333

Since m is positive : Image is virtual and erect

Since |m| < 1 : Image is diminished