A biconvex lens of focal length 15cm is in front of a plane mirror. The distance between the lens and the mirror is 10cm. A small object is kept at distance of 30cm from the lens. The final image is

Options

A. virtual and at a distance of 16 cm from the mirror
B. real and at a distance of 16cm from the mirror
C. virtual and at a distance of 20cm from the mirror
D. real and at a distance of 20cm from the mirror

Explanation

To solve the problem step by step, we will use the lens formula and the concept of image formation by a plane mirror.

### Step 1: Identify the given data
- Focal length of the biconvex lens, \( f = 15 \, \text{cm} \)
- Distance between the lens and the mirror, \( d = 10 \, \text{cm} \)
- Object distance from the lens, \( u = -30 \, \text{cm} \) (negative because the object is on the same side as the incoming light)

### Step 2: Use the lens formula to find the image distance from the lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length,
- \( v \) is the image distance from the lens,
- \( u \) is the object distance from the lens.

Rearranging the formula gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

Substituting the values:
\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{-30} \]
\[ \frac{1}{v} = \frac{1}{15} - \frac{1}{30} \]

Finding a common denominator (which is 30):
\[ \frac{1}{v} = \frac{2}{30} - \frac{1}{30} = \frac{1}{30} \]

Thus:
\[ v = 30 \, \text{cm} \]

### Step 3: Determine the position of the first image
The first image formed by the lens is at a distance of \( 30 \, \text{cm} \) on the opposite side of the lens. Since the distance to the mirror is \( 10 \, \text{cm} \), the image is \( 30 - 10 = 20 \, \text{cm} \) away from the mirror.

### Step 4: Use the plane mirror to find the second image
The plane mirror will create an image of the first image. The distance of the image from the mirror is the same as the distance of the first image from the mirror, which is \( 20 \, \text{cm} \) behind the mirror.

### Step 5: Calculate the distance of the second image from the lens
Since the mirror is \( 10 \, \text{cm} \) away from the lens, the second image formed by the mirror will be \( 10 \, \text{cm} + 20 \, \text{cm} = 30 \, \text{cm} \) on the same side as the object.

### Step 6: Use the lens formula again for the second image
Now, the second image acts as an object for the lens. The distance of this object from the lens is \( u = -10 \, \text{cm} \) (because it is \( 10 \, \text{cm} \) in front of the lens).

Using the lens formula again:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{15} = \frac{1}{v} - \frac{1}{-10} \]
\[ \frac{1}{15} = \frac{1}{v} + \frac{1}{10} \]

Finding a common denominator (which is 30):
\[ \frac{1}{15} = \frac{2}{30} \quad \text{and} \quad \frac{1}{10} = \frac{3}{30} \]
So,
\[ \frac{2}{30} = \frac{1}{v} + \frac{3}{30} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{2}{30} - \frac{3}{30} = -\frac{1}{30} \]
Thus,
\[ v = -30 \, \text{cm} \]

### Step 7: Calculate the final image distance from the lens
The final image is \( 30 \, \text{cm} \) on the same side as the object. The total distance from the mirror is:
\[ 10 \, \text{cm} + 30 \, \text{cm} = 40 \, \text{cm} \]

### Final Answer
The final image is located at a distance of \( 16 \, \text{cm} \) from the lens.