A converging lens and a diverging mirror are placed at a separation of 15 cm. The focal length of the lens is 25 cm and that of the mirror is 40 cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis ?

Explanation

To solve the problem step by step, we will follow the principles of optics involving a converging lens and a diverging mirror.

### Step 1: Understand the Setup
We have a converging lens with a focal length (f_lens) of 25 cm and a diverging mirror with a focal length (f_mirror) of -40 cm (negative because it is a diverging mirror). The distance between the lens and the mirror is 15 cm.

### Step 2: Determine the Image Position after Reflection
To have the light rays coming out parallel to the principal axis after passing through the lens, the image formed by the mirror must be at the focal point of the lens. Since the focal length of the lens is 25 cm, the image must be formed 25 cm in front of the lens.

### Step 3: Calculate the Required Object Distance from the Mirror
If the image is to be formed 25 cm in front of the lens, and the distance between the lens and the mirror is 15 cm, we can find the distance of the image from the mirror (v_mirror):
- Distance from lens to image = 25 cm
- Distance from lens to mirror = 15 cm
- Therefore, distance from mirror to image (v_mirror) = 25 cm - 15 cm = 10 cm.

### Step 4: Use the Mirror Formula
Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \(f\) = focal length of the mirror = -40 cm
- \(v\) = image distance from the mirror = 10 cm (positive, as per the sign convention)
- \(u\) = object distance from the mirror (which we need to find)

Substituting the values into the mirror formula:
\[ \frac{1}{-40} = \frac{1}{10} + \frac{1}{u} \]

### Step 5: Solve for Object Distance (u)
Rearranging the equation:
\[ \frac{1}{u} = \frac{1}{-40} - \frac{1}{10} \]
Finding a common denominator (which is 40):
\[ \frac{1}{u} = \frac{-1}{40} - \frac{4}{40} = \frac{-5}{40} \]
Thus,
\[ u = -\frac{40}{5} = -8 \text{ cm} \]
(Note that the negative sign indicates that the object is placed on the same side as the incoming light.)

### Step 6: Calculate the Position of the Object Relative to the Lens
Since the object distance from the mirror is 8 cm, we need to find the distance from the lens:
- Total distance between lens and mirror = 15 cm
- Therefore, the distance of the object from the lens = 15 cm - 8 cm = 7 cm.

### Final Answer
The point source should be placed **7 cm** from the lens.

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