Which of the following lenses would you prefer to use while reading very small letters?

A convex lens of focal length 5 cm .

A convex lens of focal length 50 cm .

A concave lens of focal length 50 cm .

A convex lens of focal length 5 cm

Find the power of a concave lens of focal length 2 m .

The focal length of a diverging lens (such as a concave lens) is negative, and it is the distance from which a collimated beam seems to be diverging after passing through the lens. Given: focal length of the lens, f = - 2 m The power of the lens, P = 1 f = 1 - 2 = - 0 .5 D .

A 20 mm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm . If the distance of the object from the lens is 15 cm , then find the magnification of the image.

Given, The height of the object, h 1 = + 2 cm Focal length, f = + 10 cm Object distance, u = - 15 cm We have the lens formula, 1 v - 1 u = 1 f ⇒ 1 v - 1 - 15 = 1 10 ⇒ 1 v + 1 15 = 1 10 ⇒ 1 v = 1 10 - 1 15 ⇒ 1 v = 15 - 10 150 = 1 30 ∴ v = 30 cm Magnification, m = v u = 30 - 15 = - 2 Thus, we can say that the position of the image is formed at a distance of 30 cm from the convex lens. As the value of the magnification is more than 1 , the image formed is larger than the object. The minus sign of magnification indicates that the image is formed below the principal axis. Thus, we can say that the image formed is real and inverted.

A 20 mm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm . If the distance of the object from the lens is 15 cm , then find the natures position and size of the image.

Given, The height of the object, h 1 = + 2 cm Focal length, f = + 10 cm Object distance, u = - 15 cm We have the lens formula, 1 v - 1 u = 1 f ⇒ 1 v - 1 - 15 = 1 10 ⇒ 1 v + 1 15 = 1 10 ⇒ 1 v = 1 10 - 1 15 ⇒ 1 v = 15 - 10 150 = 1 30 ∴ v = 30 cm Thus, we can say that the position of the image is formed at a distance of 30 cm from the convex lens. Thus, we can say that the image formed is real and inverted.

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm . Draw its ray diagram.

Object distance, u = - 25 cm Object height, h o = 5 cm Focal length, f = + 10 cm According to the lens formula, 1 v - 1 u = 1 f 1 v = 1 f + 1 u = 1 10 - 1 25 = 15 250 v = 250 15 = 16.67 cm The positive value of v shows that the image is formed at the other side of the lens.