List the sign conventions for reflection of light by spherical mirrors. Draw a diagram and apply these conventions in the determination of focal length of a spherical mirror which forms a three times magnified real image of an object placed 16 cm infront of it.

Or

State the law of refraction of light that defines the refractive index of a medium with respect to the other. Express it mathematically. How is refractive index of any medium ‘A’ with respect to a medium ‘B’ related to the speed of propagation of light in two media A and B ? State the name of this constant when one medium is vacuum or air. The refractive indices of glass and water with respect to vacuum are 3/2 and 4/3 respectively. If the speed of light in glass is $2 \times 10^8$ m/s, find the speed of light in (i) vacuum, (ii) water.

Part 1: New Cartesian Sign Conventions for Spherical Mirrors

1. Position of Object: The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.
2. Measurement of Distances: All distances parallel to the principal axis are measured from the pole ($P$) of the mirror.
3. Direction of Incident Light: Distances measured in the direction of incident light (to the right of the pole, along the $+x$-axis) are taken as positive, while those measured against the direction of incident light (to the left of the pole, along the $-x$-axis) are taken as negative.
4. Perpendicular Heights Above Axis: Distances measured perpendicular to and above the principal axis (along the $+y$-axis) are taken as positive.
5. Perpendicular Heights Below Axis: Distances measured perpendicular to and below the principal axis (along the $-y$-axis) are taken as negative.

Numerical Solution (First Option)

Given:

• Object distance, $u = -16\text{ cm}$ (by sign convention)
• Magnification, $m = -3$ (since the image is real, magnification is negative)

Using the magnification formula for spherical mirrors:

$$m = -\frac{v}{u}$$

$$-3 = -\frac{v}{-16}$$

$$-3 = \frac{v}{16} \implies v = -48\text{ cm}$$

Now, using the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

$$\frac{1}{f} = \frac{1}{-48} + \frac{1}{-16}$$

$$\frac{1}{f} = \frac{-1 - 3}{48} = \frac{-4}{48} = -\frac{1}{12}$$

$$f = -12\text{ cm}$$

Thus, the focal length of the spherical mirror is $-12\text{ cm}$ (the negative sign indicates it is a concave mirror).

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OR Option Solution

1. Law of Refraction (Snell's Law):

The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given pair of media.

Mathematically:

$$\frac{\sin i}{\sin r} = \text{constant} = n_{21}$$

where $n_{21}$ is the refractive index of medium 2 with respect to medium 1.

1. Relation with Speed of Propagation:

The refractive index of medium A with respect to medium B ($n_{AB}$) is given by the ratio of the speed of light in medium B ($v_B$) to the speed of light in medium A ($v_A$):

$$n_{AB} = \frac{v_B}{v_A}$$

1. Name of the Constant:

When one medium is vacuum or air, this constant is called the absolute refractive index of the medium.

1. Numerical Calculations:

Given:

• Refractive index of glass, $n_g = \frac{3}{2}$
• Refractive index of water, $n_w = \frac{4}{3}$
• Speed of light in glass, $v_g = 2 \times 10^8\text{ m/s}$

(i) Speed of light in vacuum ($c$):

$$n_g = \frac{c}{v_g}$$

$$\frac{3}{2} = \frac{c}{2 \times 10^8}$$

$$c = \frac{3}{2} \times 2 \times 10^8 = 3 \times 10^8\text{ m/s}$$

(ii) Speed of light in water ($v_w$):

$$n_w = \frac{c}{v_w}$$

$$\frac{4}{3} = \frac{3 \times 10^8}{v_w}$$

$$v_w = \frac{3 \times 3 \times 10^8}{4} = 2.25 \times 10^8\text{ m/s}$$