A thin rod of length `d//3` is placed along the principal axis of a concave mirror of focal length = d such that its image, which is real and elongated, just touches the rod. Find the length of the image ?

Options

A. (a)f
B. (b)`1/2f`
C. (c)2f
D. (d)`1/4f`

Explanation

To solve the problem step by step, we will use the concepts of ray optics, specifically the mirror formula and magnification.

### Step 1: Understand the Problem
We have a thin rod of length \( \frac{d}{3} \) placed along the principal axis of a concave mirror with a focal length \( f = d \). The image formed is real and elongated, and it just touches the rod.

### Step 2: Identify the Magnification Formula
The magnification \( m \) of a mirror is given by the formula:
\[ m = \frac{\text{Length of Image}}{\text{Length of Object}} \]
Let the length of the image be \( L \). The length of the object (the rod) is \( \frac{d}{3} \), so we can express the magnification as:
\[ m = \frac{L}{\frac{d}{3}} = \frac{3L}{d} \]

### Step 3: Determine the Condition for Touching
The image just touches the rod when the image distance \( v \) is equal to the object distance \( u \) plus the length of the rod:
\[ v = u + \frac{d}{3} \]

### Step 4: Use the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting \( f = d \):
\[ \frac{1}{d} = \frac{1}{v} + \frac{1}{u} \]

### Step 5: Substitute for \( v \)
From the condition \( v = u + \frac{d}{3} \), we can substitute \( v \) into the mirror formula:
\[ \frac{1}{d} = \frac{1}{u + \frac{d}{3}} + \frac{1}{u} \]

### Step 6: Solve for \( u \)
To solve for \( u \), we first find a common denominator:
\[ \frac{1}{d} = \frac{u + \frac{d}{3} + u}{u(u + \frac{d}{3})} \]
This simplifies to:
\[ \frac{1}{d} = \frac{2u + \frac{d}{3}}{u^2 + \frac{du}{3}} \]

Cross-multiplying gives:
\[ u^2 + \frac{du}{3} = d(2u + \frac{d}{3}) \]

### Step 7: Rearranging the Equation
Rearranging the equation leads to:
\[ u^2 - 2du + \frac{d^2}{3} = 0 \]

### Step 8: Use the Quadratic Formula
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{2d \pm \sqrt{(2d)^2 - 4 \cdot 1 \cdot \frac{d^2}{3}}}{2} \]
This simplifies to:
\[ u = d \pm \frac{d}{\sqrt{3}} \]

### Step 9: Find \( v \)
Now substituting \( u \) back to find \( v \):
\[ v = u + \frac{d}{3} \]

### Step 10: Calculate the Length of the Image
Using the magnification formula:
\[ m = \frac{3L}{d} \]
And substituting \( m \) we found earlier, we can solve for \( L \):
\[ L = \frac{3}{2} \cdot \frac{d}{3} = \frac{d}{2} \]

### Final Answer
The length of the image \( L \) is \( \frac{d}{2} \).

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