Here’s a lot of tough questions you could expect in the Class 12 physics paper.
Q1: A convex lens of focal length 20 cm is placed 10 cm in front of a convex mirror of radius 15 cm. Where should a point object be placed in front of the lens so that it images on itself?
Approach to the problem:
a) The question arises on how to place an object behind the mirror as the centre of curvature of a convex mirror is behind the mirror? The answer to this problem lies in the fact that the image formed by a convex lens becomes the virtual object for the mirror. For this to happen, rays after refraction will fall normally on the mirror. So, after normal incidence, rays will retrace their path (as reflected rays) and will form a final image on the object itself.
b) As the centre of curvature is 15 cm and the distance between the lens and the mirror is 10 cm so for convex lens image distance is 25 cm and focal length is 20 cm. Now by applying the lens formula
1/v – 1/u = 1/f
1/u = 1/v – 1/f
⇒ 1/u = 1/25 – 1/20
= 4 – 5 /5 × 5 × 4 = – 1 / 100
⇒ u = –100 cm
Note: In optics, normally we find a system of optical elements with the stipulation that the image is formed on the object itself. In such cases there has to be a mirror at the end of the optical system and rays have to be incident normally on the mirror in order to retrace the path.
Q2: Converging rays are intercepted by a concave lens of focal length 20 cm. If the rays originally converged to a point 10 cm behind the lens, where will they now converge after passing through the lens?
Approach to the problem:
a) Whenever converging rays fall on an optical element, the object is virtual. In the absence of an optical element, the point of convergence is the position of the virtual object and the object distance u=distance from optical element to the point of convergence behind the optical element.
b) In optics, students are advised to be clear about sign convention. Focal length of a concave lens is negative whereas focal length of a convex lens is positive.
c) As per the question, converging rays are falling so u =10 cm. Focal length f = –20 cm. Now by applying lens formula 1/v – 1/u = 1 / f
⇒ 1/ v = 1/ f + 1/ u = – 1 / 20 + 1/10
= – 1 + 2 / 20 = 1/20
⇒ v = 20 cm
Q3: Two conducting, concentric sphere have radii a and b. The outer sphere is given a charge What is the charge on inner sphere if it is earthed?
Misconception : Generally students think that when body is earthed, the charge goes to earth ie, Q = 0 on the body.
But when the object is earthed, it means its potential is zero, but charge on it may not be zero. So, in order to solve this, take potential on it as zero and assume that charge on it (inner sphere) is q´ and hence solve for q´.
V = 1/4π∈° [q´/a – q´/b + (Q + q´)/b] = 0
⇒ q´ / a – q´ / b + Q /b + q / b = 0
q´/ a = – Q /b´
⇒ q´= – Qa /b
The questions have been compiled by subject experts of physics at Studymate