The image produced by a concave mirror is one-quarter the size of object. If the object is moved 5cm closer to the mirror, the image will only be half the size of the object. The focal length of mirror is

Options

A. `f=5.0cm`
B. `f=2.5cm`
C. `f=7.5cm`
D. `f=10cm`

Explanation

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

### Step 1: Define the variables
Let the initial distance of the object from the mirror be \( u \). The magnification \( m \) is given as \( \frac{1}{4} \).

### Step 2: Relate the magnification to the image distance
The magnification \( m \) is defined as:
\[ m = -\frac{v}{u} \]
Given that \( m = \frac{1}{4} \), we can write:
\[ -\frac{v}{u} = \frac{1}{4} \]
This implies:
\[ v = -\frac{u}{4} \]

### Step 3: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting \( v \) into the mirror formula:
\[ \frac{1}{f} = \frac{1}{-\frac{u}{4}} + \frac{1}{u} \]
This simplifies to:
\[ \frac{1}{f} = -\frac{4}{u} + \frac{1}{u} = -\frac{4 - 1}{u} = -\frac{3}{u} \]
Thus, we have:
\[ f = -\frac{u}{3} \quad \text{(Equation 1)} \]

### Step 4: Object moved 5 cm closer
Now, the object is moved 5 cm closer to the mirror, so the new object distance \( u' \) is:
\[ u' = -u - 5 \]
The new magnification is \( \frac{1}{2} \), so:
\[ -\frac{v'}{u'} = \frac{1}{2} \]
This gives:
\[ v' = -\frac{u' }{2} = -\frac{-u - 5}{2} = \frac{u + 5}{2} \]

### Step 5: Apply the mirror formula again
Using the mirror formula for the new distances:
\[ \frac{1}{f} = \frac{1}{v'} + \frac{1}{u'} \]
Substituting \( v' \) and \( u' \):
\[ \frac{1}{f} = \frac{2}{u + 5} + \frac{1}{-u - 5} \]
This simplifies to:
\[ \frac{1}{f} = \frac{2}{u + 5} - \frac{1}{u + 5} = \frac{1}{u + 5} \]
Thus, we have:
\[ f = u + 5 \quad \text{(Equation 2)} \]

### Step 6: Equate the two expressions for focal length
From Equation 1 and Equation 2:
\[ -\frac{u}{3} = u + 5 \]
Multiplying through by -3 to eliminate the fraction:
\[ u = -3(u + 5) \]
Expanding gives:
\[ u = -3u - 15 \]
Combining like terms:
\[ 4u = -15 \quad \Rightarrow \quad u = -\frac{15}{4} = -3.75 \text{ cm} \]

### Step 7: Substitute \( u \) back to find \( f \)
Now substituting \( u \) back into Equation 1:
\[ f = -\frac{-3.75}{3} = 1.25 \text{ cm} \]

### Step 8: Conclusion
The focal length of the concave mirror is:
\[ f = -1.25 \text{ cm} \]
(Note: The negative sign indicates that it is a concave mirror.)