A concave spherical surface of radius of curvature 10 cm separates two medium X & Y of refractive index 4/3 & 3/2 respectively. If the object is placed along principal axis in medium X then :
A ray of light falls on a transparent sphere with centre at C as shown in fig.
The ray emerges from the sphere parallel to line AB. The refractive index of the sphere is
An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between these walls. The lens is kept at distance \(x\) in front of the second wall. The required focal length of the lens will be
For an image of equal size, the magnification is \(m = -1\), which means the image distance \(v = x\) is equal to the object distance \(u = x\). Using the lens formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{2}{x} ⇒ f = \frac{x}{2}\).
A convex lens ‘A’ of focal length 20 cm and a concave lens ‘B’ of focal length 5 cm are kept along the same axis with a distance ‘\(d\)’ between them. If a parallel beam of light falling on ‘A’ leaves ‘B’ as a parallel beam, then the distance ‘\(d\)’ in cm will be
For an incoming parallel beam to emerge parallel from the combination, the focus of the first lens must coincide with the virtual focus of the second lens. Thus, \(d = f_1 - |f_2| = 20 - 5 = 15\) cm.