A concave lens of focal length 20 cm product an image half in size of the real object. The distance of the real object is

Options

A. 20cm
B. 30 cm
C. 40 cm
D. 60 cm

Answer

To solve the problem step by step, we will use the lens formula and the magnification formula for a concave lens.

### Step 1: Identify the given values
- Focal length of the concave lens (F) = -20 cm (negative because it is a concave lens)
- Magnification (M) = Size of the image (HI) / Size of the object (HO) = 1/2 (the image is half the size of the object)

### Step 2: Use the magnification formula
The magnification for a lens is given by the formula:
\[ M = \frac{HI}{HO} = \frac{V}{U} \]

Where:
- V = image distance
- U = object distance

Given that the image is half the size of the object:
\[ M = \frac{1}{2} = \frac{V}{U} \]

From this, we can express V in terms of U:
\[ V = \frac{U}{2} \]

### Step 3: Use the lens formula
The lens formula for a lens is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]

Substituting the values we have:
\[ \frac{1}{-20} = \frac{1}{V} - \frac{1}{U} \]

### Step 4: Substitute V in the lens formula
Now, substitute \( V = \frac{U}{2} \) into the lens formula:
\[ \frac{1}{-20} = \frac{1}{\frac{U}{2}} - \frac{1}{U} \]

This simplifies to:
\[ \frac{1}{-20} = \frac{2}{U} - \frac{1}{U} \]
\[ \frac{1}{-20} = \frac{2 - 1}{U} \]
\[ \frac{1}{-20} = \frac{1}{U} \]

### Step 5: Solve for U
Taking the reciprocal of both sides gives:
\[ U = -20 \text{ cm} \]

### Step 6: Interpret the result
The negative sign indicates that the object is placed on the same side as the incoming light (which is the convention for real objects in optics).

### Final Answer:
The distance of the real object from the concave lens is 20 cm.

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