A diverging meniscus lens of 1.5 refractive index has concave surfaces of radii 3 and 4 cm. The position of image if an object is placed 12cm infront of the lens is

Options

A. `-24cm`
B. `-8cm`
C. `8cm`
D. 24cm

Explanation

To solve the problem of finding the position of the image formed by a diverging meniscus lens, we will follow these steps:

### Step 1: Identify the parameters of the lens
- Refractive index (μ) = 1.5
- Radius of curvature of the first surface (R1) = -3 cm (concave surface)
- Radius of curvature of the second surface (R2) = -4 cm (concave surface)
- Object distance (u) = -12 cm (since the object is placed in front of the lens)

### Step 2: Use the Lens Maker's Formula to find the focal length (f)
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{-3} - \frac{1}{-4} \right) \]
\[ = 0.5 \left( -\frac{1}{3} + \frac{1}{4} \right) \]
Finding a common denominator (12):
\[ = 0.5 \left( -\frac{4}{12} + \frac{3}{12} \right) \]
\[ = 0.5 \left( -\frac{1}{12} \right) = -\frac{1}{24} \]
Thus, the focal length (f) is:
\[ f = -24 \text{ cm} \]

### Step 3: Apply the Lens Formula to find the image distance (v)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-24} = \frac{1}{v} - \frac{1}{-12} \]
This simplifies to:
\[ \frac{1}{-24} = \frac{1}{v} + \frac{1}{12} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-24} - \frac{1}{12} \]
Finding a common denominator (24):
\[ \frac{1}{v} = -\frac{1}{24} - \frac{2}{24} = -\frac{3}{24} \]
Thus:
\[ \frac{1}{v} = -\frac{1}{8} \]
Therefore, the image distance (v) is:
\[ v = -8 \text{ cm} \]

### Step 4: Interpret the result
The negative sign indicates that the image is formed on the same side as the object, which is typical for a diverging lens.

### Final Answer:
The position of the image is at -8 cm.

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