When an object is placed `60 cm` in front of a diverging lens, a virtual image is formed `20 cm` from the lens. The lens is made of a refractive index `mu=1.65` and its two spherical surfaces have the same radius of curvature. What is the value of this radius?

Explanation

To solve the problem step by step, we will use the lens formula and the lens maker's formula.

### Step 1: Identify the given values
- Object distance (u) = -60 cm (negative because the object is on the same side as the incoming light)
- Image distance (v) = -20 cm (negative because it is a virtual image)
- Refractive index (μ) = 1.65

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Substituting the values of v and u:
\[ \frac{1}{f} = \frac{1}{-20} - \frac{1}{-60} \]

### Step 3: Calculate the right-hand side
Calculating the right-hand side:
\[ \frac{1}{f} = -\frac{1}{20} + \frac{1}{60} \]

Finding a common denominator (which is 60):
\[ \frac{1}{f} = -\frac{3}{60} + \frac{1}{60} = -\frac{2}{60} = -\frac{1}{30} \]

Thus, we have:
\[ f = -30 \text{ cm} \]

### Step 4: Use the lens maker's formula
The lens maker's formula for a lens with two spherical surfaces of the same radius of curvature (R) is:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Since \(R_1 = -R\) and \(R_2 = R\) for a diverging lens:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{-1}{R} - \frac{1}{R} \right) = (\mu - 1) \left( \frac{-2}{R} \right) \]

### Step 5: Substitute the values into the lens maker's formula
Substituting the values we have:
\[ \frac{1}{-30} = (1.65 - 1) \left( \frac{-2}{R} \right) \]
\[ \frac{1}{-30} = 0.65 \left( \frac{-2}{R} \right) \]

### Step 6: Solve for R
Rearranging gives:
\[ \frac{1}{-30} = -\frac{1.3}{R} \]
Multiplying both sides by -R:
\[ \frac{R}{30} = 1.3 \]
Thus,
\[ R = 1.3 \times 30 = 39 \text{ cm} \]

### Final Answer
The radius of curvature \(R\) is **39 cm**.

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