Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, $${{\sin {\theta _1}} \over {\sin {\theta _2}}} = {{{n_2}} \over {{n_1}}}$$, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation $$n = \left( {{c \over v}} \right) = \pm \,\sqrt {{\varepsilon _r}{\mu _r}} $$, where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, $$\varepsilon $$r and $$\mu$$r, are the relative permittivity and permeability of the medium, respectively.
In normal materials, both $$\varepsilon$$r and $$\mu$$r , are positive, implying positive n for the medium. When both $$\varepsilon$$r and $$\mu$$r are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating and physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.
Choose the correct statement.
A biconvex lens is formed with two planoconvex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both curved surface are of the same radius of curvature R = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be
A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is
A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as (Take g = 10 m/s$$^2$$)