Distance of an object from a concave lens of focal length 20 cm is 40 cm. Then linear magnification of the image

Options

A. 1
B. `lt1`
C. `gt1`
D. zero

Explanation

To find the linear magnification of an image formed by a concave lens, we can use the lens formula and the magnification formula. Here's the step-by-step solution:

### Step 1: Identify the given values
- Focal length of the concave lens (f) = -20 cm (the focal length is negative for concave lenses)
- Object distance (u) = -40 cm (the object distance is also negative as per the sign convention for lenses)

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) = focal length of the lens
- \( v \) = image distance
- \( u \) = object distance

Substituting the known values into the lens formula:
\[ \frac{1}{-20} = \frac{1}{v} - \frac{1}{-40} \]

### Step 3: Simplify the equation
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{-20} + \frac{1}{40} \]
Finding a common denominator (which is 40):
\[ \frac{1}{v} = \frac{-2}{40} + \frac{1}{40} = \frac{-2 + 1}{40} = \frac{-1}{40} \]

### Step 4: Calculate the image distance (v)
Taking the reciprocal gives:
\[ v = -40 \text{ cm} \]
This indicates that the image is formed on the same side as the object, which is typical for a concave lens.

### Step 5: Calculate the linear magnification (m)
The magnification (m) is given by the formula:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
Where:
- \( h' \) = height of the image
- \( h \) = height of the object

Substituting the values of \( v \) and \( u \):
\[ m = -\frac{-40}{-40} = -1 \]

### Step 6: Interpret the result
The negative sign indicates that the image is virtual and upright. The magnitude of the magnification is 1, meaning the image size is the same as the object size.

### Final Answer
The linear magnification of the image is \( -1 \).

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