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, we need to find the distance of the real object from a concave lens given its focal length and the magnification produced by the lens. Let's go through the solution step by step.

### Step 1: Understand the Given Information
- Focal length of the concave lens (f) = -20 cm (negative because it is a concave lens)
- Magnification (m) = height of image (h_i) / height of object (h_o) = 1/2 (the image is half the size of the object)

### Step 2: Use the Magnification Formula
The magnification for a lens is also given by the formula:
\[ m = \frac{v}{u} \]
where:
- v = distance of the image from the lens
- u = distance of the object from the lens

From the magnification provided:
\[ \frac{v}{u} = \frac{1}{2} \]
This implies:
\[ v = \frac{u}{2} \] (Equation 1)

### Step 3: Use the Lens Formula
The lens formula relates the object distance (u), image distance (v), and focal length (f):
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

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

### Step 4: Substitute v from Equation 1 into the Lens Formula
Now, substitute \( v = \frac{u}{2} \) into the lens formula:
\[ \frac{1}{-20} = \frac{1}{\frac{u}{2}} - \frac{1}{u} \]

### Step 5: Simplify the Equation
The left side remains as is:
\[ \frac{1}{-20} = \frac{2}{u} - \frac{1}{u} \]
This simplifies to:
\[ \frac{1}{-20} = \frac{2 - 1}{u} \]
\[ \frac{1}{-20} = \frac{1}{u} \]

### Step 6: Solve for u
Cross-multiplying gives:
\[ u = -20 \, \text{cm} \]

### Step 7: Interpret the Sign
The negative sign indicates that the object is on the same side as the incoming light (the left side of the lens). Therefore, the distance of the real object is:
\[ |u| = 20 \, \text{cm} \]

### Final Answer
The distance of the real object is **20 cm**.

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