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 convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image

Options

A. `(d)/(2)`
B. d
C. `2d`
D. `(3)/(d)`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the initial image formation by the convex lens
Given:
- Focal length of the convex lens (f₁) = 30 cm
- The object is at infinity, which means the image formed by the convex lens is at its focus.

Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Since the object is at infinity (u → ∞), the formula simplifies to:
\[ v = f \]
Thus, the image distance (v₁) is:
\[ v₁ = 30 \text{ cm} \]

### Step 2: Determine the size of the image formed by the convex lens
The size of the image (h₁) is given as 2 cm. This means that the magnification (M₁) produced by the convex lens can be calculated using the formula:
\[ M = \frac{h}{h_o} = \frac{v}{u} \]
Since the object is at infinity, we can say:
\[ M₁ = \frac{h₁}{h_o} \]
However, we do not need the object height (h₀) since we already have h₁.

### Step 3: Analyze the effect of the concave lens
Now, we place a concave lens of focal length (f₂) = -20 cm at a distance of 26 cm from the convex lens. The image formed by the convex lens acts as the object for the concave lens.

The distance of the image formed by the convex lens from the concave lens (u₂) is:
\[ u₂ = v₁ - 26 \text{ cm} = 30 \text{ cm} - 26 \text{ cm} = 4 \text{ cm} \]
Since this is on the same side as the incoming light, we take it as negative:
\[ u₂ = -4 \text{ cm} \]

### Step 4: Calculate the new image distance using the concave lens formula
Using the lens formula for the concave lens:
\[ \frac{1}{f₂} = \frac{1}{v₂} - \frac{1}{u₂} \]
Substituting the values:
\[ \frac{1}{-20} = \frac{1}{v₂} - \frac{1}{-4} \]
\[ \frac{1}{v₂} = \frac{1}{-20} + \frac{1}{4} \]
Finding a common denominator (20):
\[ \frac{1}{v₂} = -\frac{1}{20} + \frac{5}{20} = \frac{4}{20} \]
Thus,
\[ v₂ = \frac{20}{4} = 5 \text{ cm} \]

### Step 5: Calculate the new size of the image formed by the concave lens
The magnification produced by the concave lens (M₂) can be calculated using:
\[ M₂ = -\frac{v₂}{u₂} \]
Substituting the values:
\[ M₂ = -\frac{5}{-4} = \frac{5}{4} \]

The new size of the image (h₂) can be calculated as:
\[ h₂ = M₂ \times h₁ = \frac{5}{4} \times 2 \text{ cm} = 2.5 \text{ cm} \]

### Final Answer
The new size of the image is **2.5 cm**.

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