List the sign conventions that are followed in case of refraction of light through spherical lenses. Draw a diagram and apply these conventions in determining the nature and focal length of a spherical lens which forms three times magnified real image of an object placed 16 cm from the lens. OR State the laws of refraction of light. Write an expression to relate absolute refractive index of a medium with speed of light in vacuum. The refractive index of a medium ‘x’ with respect to ‘y’ is 2/3 and the refractive index of medium ‘y’ with respect to ‘z’ is 4/3. Calculate the refractive index of medium ‘z’ with respect of ‘x’.

OPTION (1): Sign conventions for refraction through spherical lenses (New Cartesian Sign Convention):

1. The object is always placed on the left side of the lens.
2. All distances are measured from the optical centre of the lens.
3. Distances measured in the direction of the incident light (left $\rightarrow$ right) are taken as POSITIVE.
4. Distances measured opposite to the direction of the incident light (right $\rightarrow$ left) are taken as NEGATIVE.
5. Heights measured above the principal axis are POSITIVE.
6. Heights measured below the principal axis are NEGATIVE.

Ray diagram (focal length determination for a converging lens):

Applying the conventions:

Given: object distance u = -16 cm (object is on left), magnification m = -3 (real, inverted image is 3$\times$).

From m = v/u: v = m $\times$ u = (-3) $\times$ (-16) = +48 cm.

Lens formula: 1/f = 1/v - 1/u = 1/48 - 1/(-16) = 1/48 + 1/16 = 1/48 + 3/48 = 4/48.

$\therefore$ f = +12 cm.

Since f is positive, the lens is a CONVEX (converging) lens of focal length 12 cm.

OPTION (2):

Laws of refraction of light:

(i) The incident ray, the refracted ray, and the normal to the refracting surface at the point of incidence all lie in the same plane.

(ii) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media (Snell's law):

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

Expression for absolute refractive index:

$$n = \frac{c}{v}$$

where c = speed of light in vacuum/vacuum and v = speed of light in the medium.

Given: n{xy} = 2/3 and n{yz} = 4/3.

We know: n{xy} = nx/ny and n{yz} = ny/nz.

Therefore, n{zx} = (nz/nx) = 1 / (nx/nz) = 1 / (n{xy} \times n_{yz}).

Or equivalently, n{xz} = n{xy} \times n_{yz} = (2/3) \times (4/3) = 8/9.

Hence, n_{zx} = 1 / (8/9) = 9/8.

The refractive index of medium z with respect to medium x = 9/8.