Question

A paper is placed in front of lens at a distance 30 cm, such that paper gets burned in minimum time. Radius of curvature of bi-convex lens is 60 cm. If refractive index of lens is \( \mu = \frac{\alpha}{10} \), then value of \( \alpha \) is?

Hint:

For bi-convex lenses, the focal length can be calculated using the lensmaker's formula. The object should be placed at the focal point for the paper to burn in minimum time.

Answer 1

Correct Answer: 15

Step 1: Formula for focal length of lens.
The focal length \( f \) of a bi-convex lens is given by the lensmaker's formula: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \( \mu \) is the refractive index, and \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces of the lens. For a bi-convex lens, \( R_1 = +60 \, \text{cm} \) and \( R_2 = -60 \, \text{cm} \), so: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{60} - \frac{1}{-60} \right) \] \[ \frac{1}{f} = (\mu - 1) \left( \frac{2}{60} \right) = \frac{(\mu - 1)}{30} \]
Step 2: Applying the condition for minimum time to burn paper.
For the paper to burn in minimum time, it must be placed at the focal point of the lens. The lens equation is: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where \( v = 30 \, \text{cm} \) (the distance of the paper from the lens) and \( u \) is the object distance (which is also equal to \( f \) for minimum time). Therefore: \[ \frac{1}{f} = \frac{1}{30} - \frac{1}{f} \] Solving for \( f \): \[ \frac{2}{f} = \frac{1}{30} \quad \Rightarrow \quad f = 60 \, \text{cm} \]
Step 3: Solving for \( \alpha \).
Now, substitute \( f = 60 \, \text{cm} \) in the lensmaker's formula: \[ \frac{1}{60} = \frac{(\mu - 1)}{30} \] Solving for \( \mu \): \[ \mu - 1 = \frac{1}{2} \quad \Rightarrow \quad \mu = \frac{3}{2} \] Since \( \mu = \frac{\alpha}{10} \), we equate: \[ \frac{3}{2} = \frac{\alpha}{10} \] Solving for \( \alpha \): \[ \alpha = 15 \]