A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and `PO=OQ`. The distance `PO`

Options

A. 5R
B. 3R
C. 2R
D. 1.5R

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have a spherical surface separating air (n1 = 1.0) and glass (n2 = 1.5). The center of curvature (C) is located in the glass. A point object P is in air, and it forms a real image Q in glass. The line PQ intersects the spherical surface at point O, and it is given that PO = OQ.

### Step 2: Define Variables
Let:
- PO = x (distance from the object P to point O)
- OQ = x (distance from point O to the image Q)
Thus, the total distance PQ = PO + OQ = x + x = 2x.

### Step 3: Apply the Refraction Formula
Using the formula for refraction at a spherical surface:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
where:
- \( n_1 = 1.0 \) (refractive index of air)
- \( n_2 = 1.5 \) (refractive index of glass)
- \( u = -2x \) (object distance, negative as per sign convention)
- \( v = x \) (image distance, positive as the image is real)

### Step 4: Substitute Values into the Formula
Substituting the values into the refraction formula:
\[ \frac{1.5}{x} - \frac{1.0}{-2x} = \frac{1.5 - 1.0}{R} \]
This simplifies to:
\[ \frac{1.5}{x} + \frac{0.5}{2x} = \frac{0.5}{R} \]
\[ \frac{1.5}{x} + \frac{0.25}{x} = \frac{0.5}{R} \]
Combining the fractions on the left gives:
\[ \frac{1.75}{x} = \frac{0.5}{R} \]

### Step 5: Rearranging to Find x
Cross-multiplying gives:
\[ 1.75R = 0.5x \]
Rearranging for x:
\[ x = \frac{1.75R}{0.5} = 3.5R \]

### Step 6: Conclusion
Since PO = x, we find that:
\[ PO = 3.5R \]

### Final Answer
Thus, the distance PO is \( 3.5R \).

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