Question

(a) The distance of an object from a concave mirror is increased from $12\,\text{cm}$ to $15\,\text{cm}$. It is observed that the new image becomes $\left(\dfrac{2}{5}\right)^{\text{th}}$ the size of the initial image. Calculate the focal length of the mirror.

(b) A light ray is incident on the horizontal upper face of a glass cube making an angle of $45^\circ$ with the face. While emerging, the ray just grazes the vertical face of the cube. Find the refractive index of the glass.

Hint:

Magnification of a concave mirror is f/(f-u); set the ratio of the two magnifications equal to 2/5. For the cube, grazing emergence means the critical angle at the side face.

Answer 1

Step 1: Understanding the Concept:
For a concave mirror, the size of the image depends on the object distance and the focal length. Magnification is \(m = \dfrac{f}{f-u}\) in the Cartesian sign convention.
Let the focal length be \(f = -F\) (negative for a concave mirror) and the object distances \(u_1 = -12\) cm, \(u_2 = -15\) cm.

Step 2: Part (a): Write the two magnifications:
\[ m_1 = \frac{f}{f - u_1} = \frac{-F}{-F + 12} = \frac{F}{F - 12}, \qquad m_2 = \frac{F}{F - 15} \]
The image size ratio is the ratio of magnifications, because the object height is the same:
\[ \frac{m_2}{m_1} = \frac{F - 12}{F - 15} = \frac{2}{5} \]

Step 3: Part (a): Solve for F:
\[ 5(F - 12) = 2(F - 15) \Rightarrow 5F - 60 = 2F - 30 \Rightarrow 3F = 30 \Rightarrow F = 10\ \text{cm} \]
Check: \(m_1 = 10/(10-12) = -5\) and \(m_2 = 10/(10-15) = -2\). The ratio is \(2/5\). Good.
So the focal length of the concave mirror is 10 cm (f = -10 cm).

Step 4: Part (b): Refraction at the top face:
The ray makes \(45^{\circ}\) with the horizontal face, so the angle of incidence (from the normal) is \(i = 45^{\circ}\). Let the angle of refraction be \(r\).
\[ \sin 45^{\circ} = n \sin r \]

Step 5: Part (b): Condition at the vertical face:
The refracted ray reaches the vertical face at an angle of incidence \(90^{\circ} - r\) (the two normals are perpendicular). Grazing emergence means this is the critical angle:
\[ \sin(90^{\circ} - r) = \cos r = \frac{1}{n} \]
Now \(\sin r = \sqrt{1 - 1/n^2}\), so \(\sin 45^{\circ} = n\sqrt{1 - 1/n^2} = \sqrt{n^2 - 1}\).
\[ \frac{1}{\sqrt{2}} = \sqrt{n^2 - 1} \Rightarrow n^2 = \frac{3}{2} \Rightarrow n = \sqrt{1.5} \approx 1.22 \]

Final Answer:
The focal length of the mirror is 10 cm, and the refractive index of the glass is about 1.22. \[ \boxed{f = -10\ \text{cm},\quad n = \sqrt{3/2} \approx 1.22} \]