A plano-convex lens has a maximum thickness of 6 cm. When placed on a horizontal table with the curved surface in contact with the table surface, then the apparent depth of the bottom most point of the lens is found to be 4 cm. If the lens is inverted such that the plane face of the lens is in contact with the surface of the table, then the apparent depth of the centre of the plane face is found to be `17/4` cm. The radius of curvature of the lens is

Options

A. 68 cm
B. 75 cm
C. 128 cm
D. 34 cm

Explanation

To solve the problem, we need to find the radius of curvature of a plano-convex lens given its maximum thickness and the apparent depths when placed in different orientations. Let's break down the solution step by step.

### Step 1: Understand the given data
- Maximum thickness of the lens (t) = 6 cm
- Apparent depth when the curved surface is down (d1) = 4 cm
- Apparent depth when the plane surface is down (d2) = 17/4 cm

### Step 2: Use the formula for refractive index
The formula for the refractive index (n) is given by:
\[ n = \frac{\text{Real Depth}}{\text{Apparent Depth}} \]
When the curved surface is in contact with the table, the real depth is the maximum thickness of the lens:
\[ n = \frac{6 \text{ cm}}{4 \text{ cm}} = \frac{3}{2} \]

### Step 3: Use the lens maker's formula
We can use the lens maker's formula in the context of the lens:
\[ \frac{n_1}{v} - \frac{n_2}{u} = \frac{n_1 - n_2}{R} \]
Where:
- \( n_1 = 1.5 \) (refractive index of the lens)
- \( n_2 = 1 \) (refractive index of air)
- \( u \) is the object distance (apparent depth when the plane surface is down)
- \( v \) is the image distance (apparent depth when the curved surface is down)

### Step 4: Substitute the values
From the second orientation:
- The apparent depth when the plane surface is down is \( d2 = \frac{17}{4} \) cm.
Thus, \( u = \frac{17}{4} \) cm and \( v = 6 \) cm.

Now substituting into the lens maker's formula:
\[ \frac{1.5}{6} - \frac{1}{\frac{17}{4}} = \frac{1.5 - 1}{R} \]

### Step 5: Simplify the equation
Calculating the left-hand side:
\[ \frac{1.5}{6} = \frac{1}{4}, \quad \text{and} \quad \frac{1}{\frac{17}{4}} = \frac{4}{17} \]
Thus, we have:
\[ \frac{1}{4} - \frac{4}{17} = \frac{1.5 - 1}{R} \]
Now, finding a common denominator for the left side:
\[ \frac{1}{4} = \frac{17}{68}, \quad \frac{4}{17} = \frac{16}{68} \]
So:
\[ \frac{17}{68} - \frac{16}{68} = \frac{1}{68} \]
Thus:
\[ \frac{1}{68} = \frac{0.5}{R} \]

### Step 6: Solve for R
Cross-multiplying gives:
\[ 0.5 \cdot 68 = R \]
Thus:
\[ R = 34 \text{ cm} \]

### Final Answer
The radius of curvature of the lens is \( R = 34 \) cm.

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