An object 4 cm in size is placed at 25 cm in front of a concave mirror of focal length 15 cm . At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the size of the image.

Given, The object size, h o = + 4 cm The object distance, u = 25 cm The focal length, f = 15 cm The image distance, v = ? The height of the image, h i = ? Using mirror formula, 1 f = 1 v + 1 u ⇒ 1 - 15 = 1 v + 1 - 25 ⇒ v = - 37 . 5 cm The magnification of the lens, m = - v u = h i h o ⇒ m = - v u = - - 37 . 5 - 25 = h i 4 Therefore, the image height h i = m × h o = - 6 cm The height of the image is 6 cm . To get a sharp image, v = 37 . 5 cm The image is real, inverted and magnified.

Where should an object be placed in front of a convex lens to get a real image of the size of the object?

At the principal focus of the lens.

Between the optical centre of the lens and its principal focus.

Draw a ray diagram and describe the image formation by a concave lens when an object is placed at (i) infinity (ii) a finite distance.

Image formation by a concave lens: When an Object is placed at Infinity: (i) When an object is placed at infinity, a virtual image is formed at the principal focus. (ii)The size of the image is much smaller than that of the object. When the object is placed anywhere on the principal axis at a finite distance: (i) First, we draw a ray parallel to the principal axis so, it appears to pass through focus after reflection. We draw another ray that passes through the optical centre so, the ray will go through without any deviation. Where both reflected rays meet is the point A ' . (ii) The image is formed between F 1 an Optical centre, The image is behind the mirror which is a real image, erect and smaller than the object which is diminished.