Physics REFRACTION AT SPHERICAL SURFACES AND BY LENSES, REFRACTION BY A LENS FOR CBSE NCERT

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color{blue}{star} REFRACTION AT SPHERICAL SURFACES AND BY LENSES
color{blue}{star} REFRACTION BY A LENS

REFRACTION AT SPHERICAL SURFACES AND BY LENSES

color{blue} ✍️ Now consider refraction at a spherical interface between two transparent media. An infinitesimal part of a spherical surface can be regarded as planar and the same laws of refraction can be applied at every point on the surface.

color{blue} ✍️ Just as for reflection by a spherical mirror, the normal at the point of incidence is perpendicular to the tangent plane to the spherical surface at that point and, therefore, passes through its centre of curvature.

color{blue} ✍️ We first consider refraction by a single spherical surface and follow it by thin lenses. A thin lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical.

color{blue} ✍️ Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, we shall obtain the lens maker’s formula and then the lens formula.

color{brown} bbul{"Refraction at a spherical surface"}

color{blue} ✍️ Figure 9.17 shows the geometry of formation of image I of an object O on the principal axis of a spherical surface with centre of curvature C, and radius of curvature R. The rays are incident from a medium of refractive index n_1, to another of refractive index n_2.

color{blue} ✍️ As before, we take the aperture (or the lateral size) of the surface to be small compared to other distances involved, so that small angle approximation can be made.

color{blue} ✍️ In particular, NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis. We have, for small angles,

color {purple}{tan ∠NOM = (MN)/(OM)}

color {purple}{tan ∠NOM = (MN)/(MC)}

color {purple}{tan ∠NIM = (MN)/(MI)}

color{blue} ✍️ Now, for ΔNOC, i is the exterior angle. Therefore, i = ∠NOM + ∠NCM

color {blue}{I = (MN)/(OM) + (MN)/(MC)}

...............(9.13)

color{blue} ✍️ Similarly, r = angle NCM - angle NIM

i.e .,

color {blue}{r = (MN)/(MC) - (MN)/(MI)}

...........(9.14)

color{blue} ✍️Now, by Snell’s law : color{brown} {n_1 sin i = n_2 sin r} or for small angles color{brown} {n_1i = n_2r}

color{blue} ✍️ Substituting i and r from Eqs. (9.13) and (9.14), we get

color {blue}{(n_1)/(OM) +n_2/(MI) = (n_2-n_1)/(MC)}

...............(9.15)

color{blue} ✍️Here, OM, MI and MC represent magnitudes of distances. Applying the Cartesian sign convention,

color{brown} {OM = –u, MI = +v, MC = +R}

color{blue} ✍️ Substituting these in Eq. (9.15), we get

color {blue}{(n_2)/v - (n_1)/u = (n_2-n_1)/R}

...............(9.16)

color{blue} ✍️ Equation (9.16) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface. It holds for any curved spherical surface.

Q 3108767608

Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed?
Class 12 Chapter 9 Example 6
Solution:

We use the relation given by Eq. (9.16). Here u = – 100 cm, v = ?, R = + 20 cm, n_1 = 1, and n_2 = 1.5.
We then have

(1.5)/v + 1/100 = 0.5/20

or v = +100 cm
The image is formed at a distance of 100 cm from the glass surface, in the direction of incident light.

REFRACTION BY A LENS

color{blue} ✍️Figure 9.18(a) shows the geometry of image formation by a double convex lens. The image formation can be seen in terms of two steps: (i) The first refracting surface forms the image I_1 of the object O [Fig. 9.18(b)]. The image I_1 acts as a virtual object for the second surface that forms the image at I [Fig. 9.18(c)]. Applying Eq. (9.15) to the first interface ABC, we get

color {blue}{(n_1)/(OB) +(n_2)/(BI_1) = (n_2-n_1)/(BC_1)}

................(9.17)

color{blue} ✍️ A similar procedure applied to the second interface* ADC gives,

color {blue}{-(n_2)/(DI_1) + (n_1)/(DI) = (n_2-n_1)/(DC_2)}

..............(9.18)

color{blue} ✍️ Note that now the refractive index of the medium on the right side of ADC is n_1
while on its left it is n_2. Further (DI_1) is negative as the distance is measuredagainst the direction of incident light.

color{blue} ✍️ For a thin lens, BI_1 = DI_1. Adding Eqs. (9.17) and (9.18), we get

color {blue}{(n_1)/(OB) + (n_1)/(DI) = (n_2-n_1) (1/((BC)_1) +1/((DC)_2) )}

.........(9.19)

color{blue} ✍️ Suppose the object is at infinity, i.e.,
OB → ∞ and DI = f, Eq. (9.19) gives

color {blue}{(n_1= (n_2-n_1) (1/(BC)_1) +1/(DC)_2)}

...............(9.20)

color{blue} ✍️ The point where image of an object placed at infinity is formed is called the focus F, of the lens and the distance f gives its focal length. A lens has two foci, F and F′, on either side of it (Fig. 9.19). By the sign convention,

color{brown} {BC_1= R_1}

color{brown} {DC_2 = - R_2}

color{blue} ✍️ So Eq. (9.20) can be written as

color {blue}{1/f = (n_21 - 1) (1/R_1- 1/R_2) \ \ \ \ ( n_(21) = (n_2)/(n_1) )}

............(9.21)

color{blue} ✍️ Equation (9.21) is known as the lens maker’s formula.

color{blue} ✍️ It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. Note that the formula is true for a concave lens also. In that case R_1 is negative, R_2 positive and therefore, f is negative. From Eqs. (9.19) and (9.20),

color{blue} ✍️ we get

color {blue}{(n_1)/(OB) + (n_1)/(DI) = (n_1)/f}

...........(9.22)

color{blue} ✍️ Again, in the thin lens approximation, B and D are both close to thenoptical centre of the lens. Applying the sign convention,

color{brown}{BO = – u, DI = +v}, we get

color {blue}{1/v - 1/u = 1/f}

...............(9.23)

color{blue} ✍️ Equation (9.23) is the familiar thin lens formula. Though we derived it for a real image formed by a convex lens, the formula is valid for both convex as well as concave lenses and for both real and virtual images.

color{blue} ✍️ It is worth mentioning that the two foci, F and F′, of a double convex or concave lens are equidistant from the optical centre. The focus on the side of the (original) source of light is called the first focal point, whereas the other is called the second focal point.

color{blue} ✍️ To find the image of an object by a lens, we can, in principle, take any two rays emanating from a point on an object; trace their paths using the laws of refraction and find the point where the refracted rays meet (or appear to meet). In practice, however, it is convenient to choose any two of the following rays:

color {blue}{(i)} A ray emanating from the object parallel to the principal axis of the lens after refraction passes through the second principal focus F′ (in a convex lens) or appears to diverge (in a concave lens) from the first principal focus F.

color {blue}{(ii)} A ray of light, passing through the optical centre of the lens, emerges without any deviation after refraction.

color {blue}{(iii)} A ray of light passing through the first principal focus (for a convex lens) or appearing to meet at it (for a concave lens) emerges parallel to the principal axis after refraction. Figures 9.19(a) and (b) illustrate these rules for a convex and a concave lens, respectively.

color{blue} ✍️ You should practice drawing similar ray diagrams for different positions of the object with respect to the lens and also verify that the lens formula, Eq. (9.23), holds good for all cases.

color{blue} ✍️ Here again it must be remembered that each point on an object gives out infinite number of rays. All these rays will pass through the same image point after refraction at the lens.

color {blue}{m = (h')/h = v/u}

..........(9.24)

color{blue} ✍️ When we apply the sign convention, we see that, for erect (and virtual) image formed by a convex or concave lens, m is positive, while for an inverted (and real) image, m is negative.
Q 3188867707

A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?
Class 12 Chapter 9 Example 7
Solution:

The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means n_1 = n_2.. This gives 1//f =0 or f → ∞. The lens in the liquid will act like a plane sheet of glass. No, the liquid is not water. It could be glycerine.

Power of a lens

color{blue} ✍️ Power of a lens is a measure of the convergence or divergence, which a lens introduces in the light falling on it. Clearly, a lens of shorter focal length bends the incident light more, while converging it in case of a convex lens and diverging it in case of a concave lens.

color{blue} ✍️ The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distant from the optical centre (Fig. 9.20).

color{brown}{tan delta= h/f : if h = 1 tan delta = 1/f" or " delta = 1/f} for small value of delta Thus

color{blue} ✍️ value of δ. Thus

color {blue}{P = 1/f}

.............(9.25)

color{blue} ✍️ The SI unit for power of a lens is dioptre (D): 1D = 1m^(–1).

color{blue} ✍️ The power of a lens of focal length of 1 metre is one dioptre. Power of a lens is positive for a converging lens and negative for a diverging lens.

color{blue} ✍️ Thus, when an optician prescribes a corrective lens of power + 2.5 D, the required lens is a convex lens of focal length + 40 cm. A lens of power of – 4.0 D means a concave lens of focal length – 25 cm.
Q 3138067802

(i) If f = 0.5 m for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive index for air-glass = 1.5.)
Class 12 Chapter 9 Example 8
Solution:

(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R_1 = +10 cm, R_2 = –15 cm.
Refractive index of air is taken as unity.
We use the lens formula of Eq. (9.22). The sign convention has to
be applied for f, R_1 and R_2.
Substituting the values, we

1/(12) = (n -1) (1/10- 1/(-15))

This gives n = 1.5.
(iii) For a glass lens in air, n_2 = 1.5, n_1 = 1, f = +20 cm. Hence, the lens formula gives

1/(20) = 0.5 ]1/(R_1) - 1/(R_2)]

For the same glass lens in water, n_2 = 1.5, n_1 = 1.33. Therefore,
(1.33)/f = (1.5 - 1.33) [1/R_1- 1/R_2]
Combining these two equations, we find f = + 78.2` cm.