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
Variation of angle of deviation δ versus angle of incidence for a prism is given the figure. The value of refractive index of prism :
A point object is moving towards a concave mirror of focal length \(25\text{ cm}\). When it is at a distance of \(20\text{ cm}\) from the mirror, its velocity is \(5\text{ cm/sec}\). Find velocity of image at that instant :-
Using the relation between object and image velocity: \(v_i = -m^2 v_o\). Here, magnification \(m = \frac{f}{f-u} =\frac{-25}{-25 - (-20)} = 5\). Hence, \(v_i = -5^2 \times 5 = -125\text{ cm/s}\). The speed is \(125\text{ cm/s}\).
Refractive index of a prism is \(cosec(A/2)\). Then minimum angle of deviation is :
Using the prism formula: \(mu = \frac{sin((A+\delta_m)/2)}{sin(A/2)}\). Substituting \(mu = cosec(A/2) =\frac{1}{sin(A/2)}\), we get \(sin((A+\delta_m)/2) = 1 ⇒ (A+\delta_m)/2 = 90^0 ⇒ \delta_m = 180^0 - A\).
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 lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since
A larger aperture lens collects more light (better visibility and brightness), has a higher resolving power (better resolution), and hence satisfies all specified criteria.
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.