For which of the pairs of u and f for a-mirror image is smaller in size

Options

A. u = - 10 cm, f = 20 cm
B. u = -20 cm, f = -30 cm
C. u = -45 cm, f = -10 cm
D. u =-60 cm, f = 30 cm

Answer

To determine which pairs of object distance (u) and focal length (f) result in a smaller mirror image size, we need to analyze the magnification for each case. The magnification (m) for a mirror is given by the formula:

\[ m = -\frac{V}{U} \]

where V is the image distance and U is the object distance. We will also use the mirror formula:

\[ \frac{1}{F} = \frac{1}{U} + \frac{1}{V} \]

which can be rearranged to find V:

\[ V = \frac{U \cdot F}{U - F} \]

Now, we will calculate the magnification for each given pair of u and f.

### Step-by-Step Solution:

**Option A:**
- Given: \( U = -10 \, \text{cm}, F = 20 \, \text{cm} \)
- Calculate V using the mirror formula:
\[ V = \frac{U \cdot F}{U - F} = \frac{-10 \cdot 20}{-10 - 20} = \frac{-200}{-30} = \frac{20}{3} \, \text{cm} \]
- Calculate magnification:
\[ m = -\frac{V}{U} = -\frac{\frac{20}{3}}{-10} = \frac{20}{30} = \frac{2}{3} \]
- Since \( m < 1 \), the image is smaller.

**Option B:**
- Given: \( U = -20 \, \text{cm}, F = -30 \, \text{cm} \)
- Calculate V:
\[ V = \frac{U \cdot F}{U - F} = \frac{-20 \cdot -30}{-20 - (-30)} = \frac{600}{10} = 60 \, \text{cm} \]
- Calculate magnification:
\[ m = -\frac{V}{U} = -\frac{60}{-20} = 3 \]
- Since \( m > 1 \), the image is larger.

**Option C:**
- Given: \( U = -45 \, \text{cm}, F = -10 \, \text{cm} \)
- Calculate V:
\[ V = \frac{U \cdot F}{U - F} = \frac{-45 \cdot -10}{-45 - (-10)} = \frac{450}{-35} = -\frac{90}{7} \, \text{cm} \]
- Calculate magnification:
\[ m = -\frac{V}{U} = -\frac{-\frac{90}{7}}{-45} = \frac{90}{315} = \frac{2}{7} \]
- Since \( m < 1 \), the image is smaller.

**Option D:**
- Given: \( U = -60 \, \text{cm}, F = 30 \, \text{cm} \)
- Calculate V:
\[ V = \frac{U \cdot F}{U - F} = \frac{-60 \cdot 30}{-60 - 30} = \frac{-1800}{-90} = 20 \, \text{cm} \]
- Calculate magnification:
\[ m = -\frac{V}{U} = -\frac{20}{-60} = \frac{1}{3} \]
- Since \( m < 1 \), the image is smaller.

### Conclusion:
The options that yield a smaller image size (magnification < 1) are A, C, and D.