A convex lens of focal length 15 cm is placed on a plane mirror. An object is placed at 30 cm from the lens. The image is

Options

A. real, at 30 cm in front of the mirror
B. real, at 30 cm behind the mirror
C. real, at 10 cm in front of the mirror
D. virtual, at 10 cm behind the mirror

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
We have a convex lens with a focal length \( f = 15 \, \text{cm} \) placed on a plane mirror. An object is placed at a distance of \( u = 30 \, \text{cm} \) from the lens.

### Step 2: Determine the focal length of the system
The focal length of the plane mirror is considered to be infinite (\( f_m = \infty \)). The effective focal length \( f \) of the combination of the lens and the mirror can be calculated using the formula for the combination of focal lengths:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_m} + \frac{1}{f_1} \]
Where \( f_1 \) is the focal length of the lens. Thus:
\[ \frac{1}{f} = \frac{1}{15} + 0 + \frac{1}{15} = \frac{2}{15} \]
So, the effective focal length \( f \) is:
\[ f = \frac{15}{2} = 7.5 \, \text{cm} \]

### Step 3: Apply the lens formula
We will use the lens formula to find the image distance \( v \):
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Substituting the known values:
\[ \frac{1}{v} - \frac{1}{30} = \frac{2}{15} \]

### Step 4: Solve for \( \frac{1}{v} \)
Rearranging the equation gives us:
\[ \frac{1}{v} = \frac{2}{15} + \frac{1}{30} \]
To add these fractions, we need a common denominator. The least common multiple of 15 and 30 is 30:
\[ \frac{1}{v} = \frac{4}{30} + \frac{1}{30} = \frac{5}{30} = \frac{1}{6} \]

### Step 5: Calculate \( v \)
Now, taking the reciprocal gives:
\[ v = 6 \, \text{cm} \]

### Step 6: Determine the nature of the image
Since \( v \) is positive, it indicates that the image is real and formed on the same side as the object.

### Conclusion
The image formed is at a distance of \( 6 \, \text{cm} \) in front of the lens (and thus in front of the mirror).