A point object is placed 30 cm in front of an equiconvex lens of radius of curvature 15 cm and made of glass of refractive index `(3)/(2)`. On placing a convex mirror of radius of curvature 15 cm behind the lens on image side, the final image is found to coincide with the object. The possible distance between convex lens and convex mirror is

Options

A. 15 cm
B. 30 cm
C. 7.5 cm
D. 45 cm

Answer

To solve the problem step by step, we will follow the principles of optics and the lens formula.

### Step 1: Determine the focal length of the lens
The formula for the focal length \( F \) of a lens is given by:

\[ \frac{1}{F} = \left( \mu - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

For an equiconvex lens:
- \( R_1 = +15 \, \text{cm} \) (convex surface)
- \( R_2 = -15 \, \text{cm} \) (the other convex surface)

Given that the refractive index \( \mu = \frac{3}{2} \):

\[ \frac{1}{F} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{15} - \left(-\frac{1}{15}\right) \right) \]

Calculating this gives:

\[ \frac{1}{F} = \frac{1}{2} \left( \frac{2}{15} \right) = \frac{1}{15} \]

Thus, the focal length \( F = 15 \, \text{cm} \).

### Step 2: Find the image distance using the lens formula
Using the lens formula:

\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \]

Where:
- \( u = -30 \, \text{cm} \) (object distance is negative in lens formula)
- \( F = 15 \, \text{cm} \)

Substituting the values:

\[ \frac{1}{v} - \left(-\frac{1}{30}\right) = \frac{1}{15} \]

This simplifies to:

\[ \frac{1}{v} + \frac{1}{30} = \frac{1}{15} \]

Rearranging gives:

\[ \frac{1}{v} = \frac{1}{15} - \frac{1}{30} \]

Finding a common denominator (30):

\[ \frac{1}{v} = \frac{2}{30} - \frac{1}{30} = \frac{1}{30} \]

Thus, \( v = 30 \, \text{cm} \). The image is formed 30 cm on the opposite side of the lens.

### Step 3: Analyze the convex mirror
The convex mirror has a radius of curvature \( R = 15 \, \text{cm} \), so its focal length \( f \) is:

\[ f = \frac{R}{2} = \frac{15}{2} = 7.5 \, \text{cm} \]

### Step 4: Determine the distance between the lens and the mirror
We want the final image to coincide with the object, which means the image formed by the convex mirror must also be at the same position as the object (30 cm from the lens).

Let \( d \) be the distance between the lens and the mirror. The image distance \( v' \) for the convex mirror can be expressed as:

\[ v' = d - 30 \, \text{cm} \]

Using the mirror formula:

\[ \frac{1}{v'} + \frac{1}{u'} = \frac{1}{f} \]

Where \( u' \) is the object distance for the mirror, which is \( -d \) (since it is virtual):

\[ \frac{1}{d - 30} - \frac{1}{d} = \frac{1}{7.5} \]

Cross-multiplying and simplifying gives:

\[ \frac{d - d + 30}{d(d - 30)} = \frac{1}{7.5} \]

This leads to:

\[ \frac{30}{d(d - 30)} = \frac{1}{7.5} \]

Cross-multiplying gives:

\[ 30 \cdot 7.5 = d(d - 30) \]

Calculating \( 30 \cdot 7.5 = 225 \):

\[ d^2 - 30d - 225 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:

\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where \( a = 1, b = -30, c = -225 \):

\[ d = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot (-225)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ d = \frac{30 \pm \sqrt{900 + 900}}{2} = \frac{30 \pm \sqrt{1800}}{2} = \frac{30 \pm 30\sqrt{2}}{2} \]

This simplifies to:

\[ d = 15 \pm 15\sqrt{2} \]

### Conclusion
The possible distances between the convex lens and convex mirror are \( 15(1 + \sqrt{2}) \, \text{cm} \) and \( 15(1 - \sqrt{2}) \, \text{cm} \), but since \( 1 - \sqrt{2} \) is negative, we only consider \( 15(1 + \sqrt{2}) \, \text{cm} \) as a valid solution.