The rear face of an equiconvex lens of focal length 30 cm is silvered that if behave like a concave mirror The focal length of mirror is [`mu_(Lens) `= 3/2]

Options

A. 7.5 cm
B. 15cm
C. 30cm
D. 20cm

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript:

### Step 1: Understand the Problem
We have an equi-convex lens with a focal length of 30 cm. The rear face of this lens is silvered, making it behave like a concave mirror. We need to find the focal length of this mirror.

### Step 2: Find the Radius of Curvature 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 equi-convex lens, both radii of curvature are equal, so we denote them as \( R \). Given:
- Focal length \( f = 30 \) cm
- Refractive index \( \mu = \frac{3}{2} \)

Substituting the values into the lens formula:
\[ \frac{1}{30} = \left(\frac{3}{2} - 1\right) \left(\frac{1}{R} - \frac{1}{(-R)}\right) \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{2} \left(\frac{2}{R}\right) \]
\[ \frac{1}{30} = \frac{1}{R} \]
Thus, the radius of curvature \( R \) is:
\[ R = 30 \text{ cm} \]

### Step 3: Determine the Focal Length of the Mirror
The rear face of the lens is silvered, and it behaves like a concave mirror. The focal length \( f_m \) of a concave mirror is given by:
\[ f_m = -\frac{R}{2} \]
Substituting the value of \( R \):
\[ f_m = -\frac{30}{2} = -15 \text{ cm} \]

### Step 4: Find the Equivalent Focal Length of the System
The system behaves like a concave mirror with the lens in front of it. The equivalent focal length \( f_{eq} \) can be found using the formula:
\[ \frac{1}{f_{eq}} = \frac{1}{f_l} - \frac{1}{f_m} + \frac{1}{f_l} \]
Where:
- \( f_l = 30 \text{ cm} \) (focal length of the lens)
- \( f_m = -15 \text{ cm} \) (focal length of the mirror)

Substituting the values:
\[ \frac{1}{f_{eq}} = \frac{1}{30} - \frac{1}{-15} + \frac{1}{30} \]
This simplifies to:
\[ \frac{1}{f_{eq}} = \frac{1}{30} + \frac{1}{15} + \frac{1}{30} \]
Finding a common denominator (which is 30):
\[ \frac{1}{f_{eq}} = \frac{1}{30} + \frac{2}{30} + \frac{1}{30} = \frac{4}{30} = \frac{2}{15} \]
Thus, the equivalent focal length \( f_{eq} \) is:
\[ f_{eq} = \frac{15}{2} = 7.5 \text{ cm} \]

### Final Answer
The focal length of the mirror is \( -7.5 \text{ cm} \) (indicating it is a concave mirror).

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