Light from a point source in air falls on a spherical glass surface. If `mu = 1.5`, and radius of curvature `= 20 cm`, the distance of light source from the glass surface is `100 cm` , at what position will the image be formed ? (NCERT Solved Example)

Options

A. `25 cm`
B. `50 cm`
C. `100 cm`
D. `200 cm`

Explanation

To solve the problem, we will use the formula for refraction at a spherical surface. The formula is given by:

\[ -\frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{r} \]

Where:
- \( n_1 \) = refractive index of the first medium (air)
- \( n_2 \) = refractive index of the second medium (glass)
- \( u \) = object distance
- \( v \) = image distance
- \( r \) = radius of curvature of the spherical surface

### Step 1: Identify the given values
- Refractive index of air, \( n_1 = 1 \)
- Refractive index of glass, \( n_2 = 1.5 \)
- Radius of curvature, \( r = 20 \, \text{cm} \)
- Object distance, \( u = -100 \, \text{cm} \) (negative as per sign convention)

### Step 2: Substitute the values into the formula
Using the formula, we substitute the known values:

\[ -\frac{1}{-100} + \frac{1.5}{v} = \frac{1.5 - 1}{20} \]

### Step 3: Simplify the equation
This simplifies to:

\[ \frac{1}{100} + \frac{1.5}{v} = \frac{0.5}{20} \]

### Step 4: Calculate the right-hand side
Calculating the right-hand side:

\[ \frac{0.5}{20} = 0.025 \]

So the equation now looks like:

\[ \frac{1}{100} + \frac{1.5}{v} = 0.025 \]

### Step 5: Isolate \(\frac{1.5}{v}\)
Now, we can isolate \(\frac{1.5}{v}\):

\[ \frac{1.5}{v} = 0.025 - \frac{1}{100} \]

Calculating \(\frac{1}{100}\):

\[ \frac{1}{100} = 0.01 \]

So now we have:

\[ \frac{1.5}{v} = 0.025 - 0.01 = 0.015 \]

### Step 6: Solve for \(v\)
Now, we can solve for \(v\):

\[ v = \frac{1.5}{0.015} \]

Calculating this gives:

\[ v = 100 \, \text{cm} \]

### Conclusion
The position of the image formed is at a distance of \(100 \, \text{cm}\) from the glass surface.

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