A point object is placed at a distance of 10 cm and its real image is formed at a distance of 20 cm from a concave mirror. If the object is moved by 0.1 cm towards the mirror, the image will shift by about

Options

A. 0.4 cm away from the mirror
B. 0.4 cm towards the mirror
C. 0.8 cm away from the mirror
D. 0.8 cm towards the mirror

Answer

To solve the problem step by step, we will use the mirror formula and analyze the situation before and after the object is moved.

### Step 1: Identify the given values
- Initial object distance (u) = -10 cm (since it's a concave mirror, we take it as negative)
- Initial image distance (v) = -20 cm (real image, hence negative)

### Step 2: Use the mirror formula to find the focal length (f)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{-10} + \frac{1}{-20} \]
Calculating the right side:
\[ \frac{1}{f} = -\frac{1}{10} - \frac{1}{20} = -\frac{2}{20} - \frac{1}{20} = -\frac{3}{20} \]
Thus, the focal length (f) is:
\[ f = -\frac{20}{3} \text{ cm} \]

### Step 3: Move the object towards the mirror
The object is moved 0.1 cm towards the mirror, so the new object distance (u') is:
\[ u' = -10 \text{ cm} + 0.1 \text{ cm} = -9.9 \text{ cm} \]

### Step 4: Use the mirror formula again to find the new image distance (v')
Using the same mirror formula:
\[ \frac{1}{f} = \frac{1}{u'} + \frac{1}{v'} \]
Substituting the known values:
\[ \frac{1}{-\frac{20}{3}} = \frac{1}{-9.9} + \frac{1}{v'} \]
This simplifies to:
\[ -\frac{3}{20} = -\frac{1}{9.9} + \frac{1}{v'} \]
Rearranging gives:
\[ \frac{1}{v'} = -\frac{3}{20} + \frac{1}{9.9} \]

### Step 5: Calculate the right side
To combine the fractions, we find a common denominator:
\[ \frac{1}{v'} = -\frac{3 \cdot 9.9}{20 \cdot 9.9} + \frac{20}{20 \cdot 9.9} \]
Calculating:
\[ \frac{1}{v'} = -\frac{29.7}{198} + \frac{20}{198} = \frac{-29.7 + 20}{198} = \frac{-9.7}{198} \]
Thus,
\[ v' = -\frac{198}{9.7} \approx -20.41 \text{ cm} \]

### Step 6: Calculate the shift in image position
The initial image position was at -20 cm, and the new image position is approximately -20.41 cm. The shift in the image position is:
\[ \text{Shift} = |v'| - |v| = 20.41 \text{ cm} - 20 \text{ cm} = 0.41 \text{ cm} \]
Since the image moves away from the mirror, the shift is approximately 0.41 cm.

### Final Answer
The image will shift by about **0.41 cm** away from the mirror.

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