Two thin lenses are in contact with each other. One of these is a concave lens of focal length 20 cm and the other is a convex lens of the same focal length. Determine the focal length of the combination and state how it behaves.

Given: The focal length of the concave lens f 1 = - 20 cm The focal length of the convex lens f 2 = + 20 cm The combined focal length f of the two thin lenses with focal lengths f 1 and f 2 is given by: 1 f = 1 f 1 + 1 f 2 Hence, the combined focal length of the given lenses: 1 f = - 1 20 + 1 20 = 0 f = ∞ Hence, the combination of the two lenses will behave like a plane glass plate.

A biconvex lens is to be made from glass of refractive index 1 . 55 , with both faces having the same radius of curvature. Determine this radius of curvature if the focal length is 20 cm .

Lens maker's formula: 1 f = μ - 1 1 R 1 - 1 R 2 where R 1 and R 2 are radii of curvature of the two faces. The focal length of the lens f = + 20 cm The refractive index of glass μ = 1 . 55 Let the radius of curvature of each face (in magnitude) = R cm Hence: 1 20 = 1 . 55 - 1 2 R ⇒ R = 22 cm The magnitude of the radius of curvature is 22 cm .

Use the lens equation and prove that for a convex lens when the object is placed within the focal length, the image will be virtual.

Virtual image is always erect. For the formation of the virtual image, the magnification produced by a lens m = v u must be a positive quantity. Let the distance of the object from the given convex lens of focal length f be equal to x . Using lens equation (putting the values with sign using the cartesian coordinate system) we can write: 1 f = 1 v + 1 x ⇒ v = f x x - f = y let for x < f (object is placed within the focal length of the lens) the value for y will be negative. Hence, the magnification by the lens m = - y - x = y x will be a positive quantity, and the image formed will be virtual.