The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convax lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image.

Options

A. `1.25 cm`
B. `2.5 cm`
C. `1.05 cm`
D. `2 cm`

Explanation

To solve the problem step by step, we will analyze the situation involving a convex lens and a concave lens.

### Step 1: Understand the initial conditions
- We have a convex lens with a focal length \( f_1 = 30 \, \text{cm} \).
- An object is at infinity, which means the rays coming from the object are parallel.
- The image formed by the convex lens is given as \( h_i = 2 \, \text{cm} \).

### Step 2: Determine the position of the image formed by the convex lens
For an object at infinity, the image formed by a convex lens is at its focus. Therefore, the image distance \( v_1 \) from the convex lens is:
\[ v_1 = f_1 = 30 \, \text{cm} \]

### Step 3: Identify the position of the concave lens
The concave lens has a focal length \( f_2 = -20 \, \text{cm} \) (negative because it is a concave lens). The distance between the convex lens and the concave lens is \( 26 \, \text{cm} \). Thus, the distance from the convex lens to the image formed by it is:
\[ d = v_1 - 26 \, \text{cm} = 30 \, \text{cm} - 26 \, \text{cm} = 4 \, \text{cm} \]
This means that the image formed by the convex lens acts as a virtual object for the concave lens, located \( 4 \, \text{cm} \) in front of it.

### Step 4: Calculate the image distance for the concave lens
Using the lens formula for the concave lens:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f = -20 \, \text{cm} \) (focal length of the concave lens)
- \( u = -4 \, \text{cm} \) (object distance for the concave lens, negative because the object is virtual and on the same side as the incoming light)

Substituting the values:
\[ \frac{1}{-20} = \frac{1}{v} - \frac{1}{-4} \]
This simplifies to:
\[ \frac{1}{v} = \frac{1}{-20} + \frac{1}{4} \]
Finding a common denominator (20):
\[ \frac{1}{v} = \frac{-1 + 5}{20} = \frac{4}{20} = \frac{1}{5} \]
Thus, the image distance \( v \) is:
\[ v = 5 \, \text{cm} \]

### Step 5: Calculate the magnification produced by the concave lens
The magnification \( m \) produced by the concave lens is given by:
\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
Where \( h_o = 2 \, \text{cm} \) (height of the original image) and \( u = -4 \, \text{cm} \):
\[ m = \frac{5}{-4} = -1.25 \]

### Step 6: Calculate the new size of the image
The new height of the image \( h_i' \) can be calculated as:
\[ h_i' = m \cdot h_o = -1.25 \cdot 2 \, \text{cm} = -2.5 \, \text{cm} \]
The negative sign indicates that the image is inverted.

### Final Answer
The new size of the image is \( 2.5 \, \text{cm} \) (inverted).

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