Question

Read the following paragraph and answer the questions.

A lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Considering image formation by a single spherical surface successively at the two surfaces of a lens, the lens maker's formula is obtained. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. This formula helps us obtain a relation between u, v and f for a lens. Lenses form images of objects and they are used in a number of optical devices, for example, microscopes and telescopes.

1. An object AB is kept in front of a composite convex lens, as shown in the figure. Will the lens produce one image? If not, explain.
   
2. A real image of an object formed by a convex lens is observed on a screen. If the screen is removed, will the image still be formed? Explain.
3. A double convex lens is made of glass with a refractive index of 1.55 with both faces of the same radius of curvature. Find the radius of curvature required if the focal length is 20 cm.
   OR
   Two convex lenses A and B of focal lengths 15 cm and 10 cm respectively are placed coaxially 'd' distance apart. A point object is kept at a distance of 30 cm in front of lens A. Find the value of 'd' so that the rays emerging from lens B are parallel to its principal axis.

Answer 1

1. Because the lens is composed of two refractive indexes, n₁ and n₂, it produces two pictures rather than one. Because the refractive index bends light, there are two refractive indexes, hence the light will bend in two directions. As a result, two distinct images will be generated.
2. Yes, just like an object, the true image is still situated at the same location in space. However, each of these objects' points will only emit light into a cone. Therefore, you wouldn't notice anything if your eye was outside of these cones.
3. We know `1/f = (mu - 1)(1/R_1 - 1/R_2)`
   Given f = 20 cm, μ = 1.55, R₁ = R, R₂ = -R
   `1/20 = (1.55 - 1)(1/R - 1/((-R)))`
   `1/20 = 0.55 xx 2/R`
   R = 22 cm
   OR
   Condition for the emergent beam to remain parallel after refraction through both, the lens in terms of their foci.
   i.e., foci of both lense should coincide.
   We have
   `1/F = 1/f_1 + 1/f_2 - d/(f_1f_2)`
   As a ray emerging from lens B one parallel to its principle,
   Therefore, the object is kept in focus, i.e.,
   F = u = -30 cm
   `1/(-30) = 1/15 + 1/10 - d/(15 xx 10)`
   `d/150 = 1/15 + 1/10 + 1/30 = (2 + 3 + 1)/30`
   d = `150/5`
   d = 30 cm
   So, d should be 30 cm.