R is the radius of curvature for both the surfaces of lens shown in figure. Refractive index `mu_(1), mu_(2) and mu_(3)` are indicated in the figure. Light rays are incident from the side of refractive index `mu_(1)` as shown. If f is focal length of lens then find `f//R` for ` mu_(1) = 1, mu_(2)= 1.5, mu_(3)= 2`.
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Explanation

To find the relationship between the focal length \( f \) and the radius of curvature \( R \) for the given lens using the Lensmaker's formula, we can follow these steps:

### Step 1: Apply the Lensmaker's Formula for Surface 1
For the first surface of the lens, we can use the Lensmaker's formula:

\[ \frac{\mu_2}{v_1} - \frac{\mu_1}{u_1} = \frac{\mu_2 - \mu_1}{R} \]

Where:
- \( \mu_1 = 1 \) (refractive index of the medium from which light is coming)
- \( \mu_2 = 1.5 \) (refractive index of the lens)
- \( u_1 = -\infty \) (object distance for parallel rays)
- \( v_1 \) is the image distance from surface 1.

Since \( u_1 \) is very large (for parallel rays), we can approximate:

\[ \frac{\mu_2}{v_1} = \frac{\mu_2 - \mu_1}{R} \]

Rearranging gives us:

\[ v_1 = \frac{\mu_2 R}{\mu_2 - \mu_1} \]

Substituting the values:

\[ v_1 = \frac{1.5 R}{1.5 - 1} = \frac{1.5 R}{0.5} = 3R \]

### Step 2: Apply the Lensmaker's Formula for Surface 2
Now, we apply the Lensmaker's formula for the second surface:

\[ \frac{\mu_3}{v_2} - \frac{\mu_2}{v_1} = \frac{\mu_3 - \mu_2}{R} \]

Where:
- \( \mu_3 = 2 \) (refractive index of the medium on the other side of the lens)
- \( v_2 \) is the final image distance from surface 2.

Substituting \( v_1 = 3R \):

\[ \frac{2}{v_2} - \frac{1.5}{3R} = \frac{2 - 1.5}{R} \]

This simplifies to:

\[ \frac{2}{v_2} - \frac{0.5}{3R} = \frac{0.5}{R} \]

### Step 3: Solve for \( v_2 \)
Rearranging gives us:

\[ \frac{2}{v_2} = \frac{0.5}{R} + \frac{0.5}{3R} \]

Finding a common denominator for the right side:

\[ \frac{2}{v_2} = \frac{0.5 \cdot 3 + 0.5}{3R} = \frac{1.5}{3R} = \frac{0.5}{R} \]

Thus:

\[ \frac{2}{v_2} = \frac{0.5}{R} \]

Cross-multiplying gives:

\[ 2R = 0.5 v_2 \implies v_2 = \frac{2R}{0.5} = 4R \]

### Step 4: Relate \( f \) to \( R \)
Since \( v_2 \) is the focal length \( f \):

\[ f = 4R \]

### Step 5: Find \( \frac{f}{R} \)
Now, we can find \( \frac{f}{R} \):

\[ \frac{f}{R} = \frac{4R}{R} = 4 \]

### Final Answer:
Thus, the value of \( \frac{f}{R} \) is \( 4 \).

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