A concave lens has a focal length of 15 cm . At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens? What is the nature of the image?

Given: The image distance, v = - 10 cm The focal length, f = - 15 cm From the lens formula, 1 f = 1 v - 1 u ⇒ 1 - 15 = 1 - 10 - 1 u ⇒ 1 u = - 1 30 ⇒ u = - 30 cm The object should be placed 30 cm from the concave lens. Magnification of a lens is defined as the ratio of the height of an image to the height of an object. It is also given in terms of image distance and object distance. It is equal to the ratio of image distance to that of object distance. The magnification, m = v u = - 10 - 30 = + 0 . 33 The magnification produced by the lens is + 0 . 33 .

An object of size 70 mm is place at 27 cm in front of a concave mirror of focal length 18 cm . At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of the image.

Height of the object h 1 = 70 mm = 7 cm Object distance u = - 27 cm The focal length of a concave lens f = - 18 cm Let the image distance be found be v . According to the mirror formula, 1 f = 1 v + 1 u ∴ 1 v = 1 f - 1 u ∴ 1 v = 1 - 18 - 1 - 27 ∴ 1 v = 9 - 18 × 27 v = - 54 cm The screen should be placed 54 cm to the left of the pole to get a sharp image. Let the height of the image be found be h 2 . Magnification M is given by, M = - v u = h 2 h 1 ∴ - - 54 - 27 = h 2 7 ∴ h 2 = - 14 cm The height of the image is - 14 cm . As the image height and image distance are negative, The image is real, inverted, and enlarged.

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

A concave lens has focal length of 15 cm . At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification and nature of the image produced by the lens.

Given: The image distance, v = - 10 c m The focal length , f = - 15 c m From the lens formula, undefined ⇒ 1 u = 1 - 10 + 1 15 = - 3 + 2 30 = - 1 30 ⇒ u = - 30 c m The object should be placed at 30 cm from the concave lens. The magnification, m = v u = - 10 - 30 = 1 3 = + 0.33 The magnification produced by the lens is + 0.33 The image formed by a concave lens is always erect and virtual.

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.