1. Refraction through a prism is show in figure. After suffering refraction at two faces of a prism, the emergent ray is always found to bend towards the base of the prism. It is observed, that \[\angle A = \angle r_1 + \angle r_2\]and \[\angle A + \angle \delta = \angle i + \angle e\]Angle between the incident ray and the emergent ray i.e., $ \angle \delta $ is known as the angle of deviation. Its value depends upon the angle of incidence, refractive index of prism material and the angle of prism.
2. When refracted ray passes symmetrically through a prism i.e., when $ r_1 $ = $ r_2 $ and i = e, the light rays undergoes minimum deviation Dm and in such an eventuality, \[n_{21} = \frac{sin\frac{A+D_m}{2}}{sin{\frac{A}{2}}}\]where $ n_{21} $ is the refractive index of prism material with respect to the medium outside.
3. For a prism of small angle (i.e., if $ \angle A $ is small enough), the angle of deviation is given by \[\delta , D = (n_{21}-1)A\]
4. Dispersion is the phenomenon of splitting of light into its component colours (or wavelengths) on passing through a dispersive medium. The pattern of colour components of light is called its spectrum. For sunlight, the spectrum consists of seven constituent colours given by the acronym VIBGYOR. In white light spectrum the violet ray is deviated the most and the red ray is deviated the most and the red ray is deviated the least.
5. Cause of dispersion in variation of refractive index with wavelength of light. In fact, \[n = A + \frac{B}{\lambda ^2}\]where A and B are two constants are a given material. As a result, the refractive index of prism and consequently the angle of deviation is maximum for violet colour ray and least for red colour ray. It results in dispersion.
6. Angular dispersion produced by a prism for white light is difference in the angles of deviation of two extreme colours i.e., violet and red colours. Mathematically, Angular dispersion = $ \delta_v - \delta_r = (n_v - n_r)A $.
7. The light, while passing through earth’s atmosphere, gets scattered by the atmospheric particles. According to Rayleigh’s law of scattering, for scattering from tiny scattering objects e.g., air molecules the intensity of the light corresponding to a wavelength in the scattered light varies inversely as the fourth power of the wavelength. Mathematically, Amount of scattering $ \propto \frac{1}{\lambda^4} $
8. Blue colour of sky, blue colour of ocean water, reddish appearance of Sun at sunrise or sunset are some common phenomenon based on Rayleigh’s scattering. Due to this very reason, red light is used in danger signals.
9. Rainbow is an example of dispersion of light, caused by tiny water droplets hanging in the atmosphere after the rains.
10. The human eye is one of the most valuable and sensitive sense organ, the human being have. Our eyes have a lens system which focus the light rays coming from an object on the retina. Retina contains rods and cone which sense light intensity and colour respectively. Retina transmits electrical signals via the optic nerve to the brain, which analyses the information received and perceives the object.
11. The eyelens has the power of accommodation t adjust it focal length so as to focus objects situated at different distance form eye at the retina.
12. The least distance of distinct vision or near point of an eye is the minimum distance from the eye at which object can be seen distinctly. For a young adult with normal vision near point is at 25 cm.
13. The farthest point up to which an eye can see objects clearly is called the far point of eye. For a normal vision, the far point of eye lies at infinity. In this situation, our eye is least strained.
14. There are four common defects of vision. These are (i) myopia or short-sightedness (ii) hypermetropia or long-sightedness (iii) presbyopia and (iv) astigmatism.
15. A myopic eye can see near objects clearly but cannot see far off objects clearly i.e., the far point of defective eye is not at infinity but has shifted nearer to the eye. This defect may arise either due to (a) excessive curvature of the cornea, or (b) elongation of eyeball. The defect can be corrected by use of a concave (diverging) lens of appropriate power.
16. In hypermetropia, a person can see distance objects clearly but cannot see nearby objects distinctly i.e., for defective eye the near point has shifted away from the eye. This defect arises either due to less curvature of cornea or contraction of the eyeball. The defect can be corrected by use of convex (converging) lens of appropriate power. With increase in age the ciliary muscles gradually weaken and power of accommodation of eye decreases. It is called presbyopia. It can be corrected by using a converging lens for reading .
17. In astigmatism, a person cannot focus simultaneously on both horizontal and vertical lines. It arises when the cornea is not spherical in shape. The problem can be rectified by using cylindrical lens of desired radius of curvature with an appropriately directed axis.
18. A microscope is used for observing magnified images of nearby tiny objects. A simple magnifier or microscope is a convex lens of small focal length held near the object such that $ u \leq f $.
19. In a simple microscope if image is formed at near point, the angular magnification of image is $ m = (1 + \frac{D}{f} ) $. However, if image is formed at infinity then magnification $ m = \frac{D}{f} $.
20. A compound microscope consists of two convex lenses, an objective lens of very small focal length ($ f_0 $) and small aperture and an eye lens of small focal length ($ f_e $) and slightly greater aperture, placed coaxially at a suitable fixed distance of distinct vision (D = 25 cm) from the eye and is virtual, inverted and highly magnified.
21. The angular magnification of a microscope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye when seen directly. Angular magnification of a compound microscope is given by :
(a) If final image is formed at near point of eye, then \[m = m_0 \times m_e = -\frac{v_0}{u_0}(1+\frac{D}{f_e})=-\frac{L}{f_0}(1+\frac{D}{f_e})\]
(b) If final image in a microscope is formed at infinity, then \[m =-\frac{L}{f_0}\frac{D}{f_e}\]
22. The resolving power of a compound microscope is its ability to show as distinct (separate), the images of two point objects lying close to each other. The limit of resolution of a microscope is measured by the minimum distance d between two point objects, whose images in microscope are seen as just separate. It is found that \[d = \frac{1.22 \lambda}{2nsin\alpha} = \frac{0.16 \lambda}{N.A.}\]where n = refractive index of medium between the object and the objective lens, $ 2\alpha $ = angle subtended by the diameter of objective lens at the focus point and N.A. = $ n sin \alpha $ = numerical aperture of objective. Resolving power of a microscope is the reciprocal of its limit of resolution. For higher resolving power the numerical aperture the numerical aperture of objective lens of microscope should be large and wavelength of light used should be as small as possible.
23. An astronomical telescope is used to form magnified and distinct images of heavenly bodies like planets stars, moons, galaxies etc. A refracting type astronomical telescope consists of a convex objective lens of large focal length and large aperture and another convex eyepiece lens of small total length and small aperture. Final image formed is inverted, magnified and at infinitly in normal adjustment.
24. The angular magnification of a telescope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object directly. It is found that in normal adjustment \[m = -\frac{f_0}{f_e}\]and length of telescope tube $ L = f_0 + f_e $.
25. In a reflecting type telescope we use a concave mirror (generally parabolic) of large aperture and large focal length as the objective and a convex lens of small focal length and aperture as the aberrations, are cheap, easy to construct and handle.
26. The limit of resolution of a telescope is measured by the angle ($ \Delta \theta $) subtended at its objective, by those two distant objects whose images are just seen separate through the telescope.
Resolving power of telescope = $ \frac{1}{\Delta \theta}= \frac{A}{1.22 \lambda} $, where A is the aperture size of the telescope objective.
