(a) Draw a labelled ray diagram to show the path of a ray of light incident obliquely on one face of a glass slab.

(b) Calculate the refractive index of the material of a glass slab. Given that the speed of light through the glass slab is $2 \times 10^8 \text{ m/s}$ and in air is $3 \times 10^8 \text{ m/s}$.

(c) Calculate the focal length of a lens, if its power is $-2.5 \text{ D}$.

OR

(a) Calculation of object distance:

1. Determine the focal length ($f$) of the lens:

The power of the corrective lens is $P = -2.5 \text{ D}$.

$$f = \frac{1}{P} = \frac{1}{-2.5} \text{ m} = -0.4 \text{ m} = -40 \text{ cm}$$

Since the focal length is negative, the lens is a concave lens.

1. Identify the image distance ($v$):

A concave lens always forms a virtual, erect image on the same side of the lens as the object. Therefore, by the Cartesian sign convention, the image distance is negative:

$$v = -10 \text{ cm}$$

1. Apply the Lens Formula:

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

Substitute the values of $f = -40 \text{ cm}$ and $v = -10 \text{ cm}$:

$$\frac{1}{-40} = \frac{1}{-10} - \frac{1}{u}$$

Rearranging to solve for $\frac{1}{u}$:

$$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-40}$$

$$\frac{1}{u} = -\frac{1}{10} + \frac{1}{40}$$

$$\frac{1}{u} = \frac{-4 + 1}{40} = -\frac{3}{40}$$

$$u = -\frac{40}{3} \text{ cm} \approx -13.33 \text{ cm}$$

Therefore, the student should place the object at a distance of $13.33 \text{ cm}$ in front of the concave lens.

(b) Ray diagram showing the position and nature of the image:

• Nature of the image: Virtual, erect, and diminished.
• Position of the image: At a distance of $10 \text{ cm}$ on the same side of the lens as the object (between the optical centre $O$ and the principal focus $F_1$).