A convex lens of focal length 30 cm forms an image of height 2 cm for an object situated at infinity. If a concave lens of focal length 20 cm is placed coaxially at a distance of 26 cm from convex lens then size of image would be

Options

A. `2.5 cm`
B. `5.0 cm`
C. `1.25 cm`
D. `4 cm`

Answer

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

### Step 1: Understand the setup
We have a convex lens with a focal length (f1) of +30 cm. An object is placed at infinity, which means that the rays of light coming from the object are parallel. The convex lens will focus these rays at its focal point.

### Step 2: Determine the position of the image formed by the convex lens
Since the object is at infinity, the image formed by the convex lens will be at its focal point, which is 30 cm from the lens. The height of the image (h1) formed by the convex lens is given as 2 cm.

### Step 3: Identify the position of the concave lens
The concave lens has a focal length (f2) of -20 cm and is placed 26 cm away from the convex lens. The image formed by the convex lens (which acts as the object for the concave lens) is located at a distance of 30 cm from the convex lens. Therefore, the distance of this image from the concave lens is:
\[ d = 30 \text{ cm} - 26 \text{ cm} = 4 \text{ cm } \]
This distance is positive because the image formed by the convex lens is on the same side as the incoming light for the concave lens.

### Step 4: Use the lens formula for the concave lens
The lens formula is given by:
\[ \frac{1}{V} - \frac{1}{U} = \frac{1}{F} \]
Where:
- \( V \) is the image distance from the concave lens,
- \( U \) is the object distance from the concave lens (which we found to be +4 cm),
- \( F \) is the focal length of the concave lens (-20 cm).

Substituting the values into the lens formula:
\[ \frac{1}{V} - \frac{1}{4} = \frac{1}{-20} \]
Rearranging gives:
\[ \frac{1}{V} = \frac{1}{4} - \frac{1}{20} \]

### Step 5: Find a common denominator and solve for V
The common denominator for 4 and 20 is 20:
\[ \frac{1}{V} = \frac{5}{20} - \frac{1}{20} = \frac{4}{20} \]
Thus:
\[ V = \frac{20}{4} = 5 \text{ cm} \]
This means the image formed by the concave lens is located 5 cm from the concave lens on the same side as the object.

### Step 6: Calculate the magnification
The magnification (m) of a lens is given by:
\[ m = \frac{h_i}{h_o} = \frac{V}{U} \]
Where:
- \( h_i \) is the height of the image,
- \( h_o \) is the height of the object (which is 2 cm for the convex lens image),
- \( V \) is the image distance (5 cm),
- \( U \) is the object distance (4 cm).

Now substituting the values:
\[ m = \frac{h_i}{2} = \frac{5}{4} \]
Thus:
\[ h_i = 2 \times \frac{5}{4} = \frac{10}{4} = 2.5 \text{ cm} \]

### Final Answer
The size of the final image formed by the concave lens is **2.5 cm**.

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