There is a thin double convex lens having surfaces of radius of curvature R and 2R. Lens is kept in air and refractive index of the material of the lens is `mu`. Focal length of the lens

Options

A. will depend on which surface light is incident
B. will not depend on which surface light is incident
C. will be equal to `(2R)/(3(mu -1))`
D. will be equal to `(2R)/(2(mu -1))`

Explanation

To find the focal length of a thin double convex lens with given radii of curvature, we can use the lens maker's formula. Here’s a step-by-step solution:

### Step 1: Identify the parameters
- Let the radius of curvature of the first surface (R1) be \( R \).
- Let the radius of curvature of the second surface (R2) be \( 2R \).
- The refractive index of the lens material is \( \mu \).

### Step 2: Assign signs to the radii of curvature
- For a double convex lens, the first surface (R1) is convex, so it is positive:
\[ R_1 = R \]
- The second surface (R2) is also convex, but since it is on the opposite side, it is considered negative:
\[ R_2 = -2R \]

### Step 3: Apply the lens maker's formula
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values of \( R_1 \) and \( R_2 \):
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-2R} \right) \]

### Step 4: Simplify the expression
Calculating the right-hand side:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{2R} \right) \]
\[ = (\mu - 1) \left( \frac{2}{2R} + \frac{1}{2R} \right) \]
\[ = (\mu - 1) \left( \frac{3}{2R} \right) \]

### Step 5: Solve for the focal length
Now, we can express the focal length \( f \):
\[ \frac{1}{f} = \frac{3(\mu - 1)}{2R} \]
Taking the reciprocal gives:
\[ f = \frac{2R}{3(\mu - 1)} \]

### Final Answer
Thus, the focal length of the lens is:
\[ f = \frac{2R}{3(\mu - 1)} \]

---