A diverging lens with magnitude of focal length `25 cm` is placed at a distance of `15 cm` from a converging lens of magnitude of focal length `20 cm`. A beam of parallel light falls on the diverging lens. The final image formed is.

Options

A. real and 6 cm from convergent lens
B. real and 40 cm from convergent lens
C. virtual and 40 cm from convergent lens
D. real and 40 cm from divergent lens

Answer

To solve the problem step by step, we will analyze the situation involving the diverging lens and the converging lens.

### Step 1: Understand the setup
We have two lenses:
- A diverging lens (concave lens) with a focal length \( f_1 = -25 \, \text{cm} \) (negative because it's a diverging lens).
- A converging lens (convex lens) with a focal length \( f_2 = +20 \, \text{cm} \).
- The distance between the two lenses is \( d = 15 \, \text{cm} \).

A beam of parallel light rays is incident on the diverging lens.

### Step 2: Find the image formed by the diverging lens
Since the light rays are parallel, they can be considered to be coming from infinity. For a diverging lens, the formula for the lens is given by:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Here, \( u \) (object distance) is taken as \( -\infty \) (since the object is at infinity), and \( f_1 = -25 \, \text{cm} \).

Using the lens formula:

\[ \frac{1}{-25} = \frac{1}{v_1} - \frac{1}{-\infty} \]

This simplifies to:

\[ \frac{1}{v_1} = \frac{1}{-25} \]

Thus,

\[ v_1 = -25 \, \text{cm} \]

This means the image formed by the diverging lens is virtual and located \( 25 \, \text{cm} \) on the same side as the object (to the left of the lens).

### Step 3: Determine the object distance for the converging lens
The virtual image formed by the diverging lens acts as an object for the converging lens. The distance of this virtual image from the converging lens is:

\[ u_2 = d + |v_1| = 15 \, \text{cm} + 25 \, \text{cm} = 40 \, \text{cm} \]

Since this is a real object for the converging lens, we take \( u_2 = -40 \, \text{cm} \).

### Step 4: Find the image formed by the converging lens
Now we apply the lens formula for the converging lens:

\[ \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \]

Substituting the values:

\[ \frac{1}{20} = \frac{1}{v_2} - \frac{1}{-40} \]

This simplifies to:

\[ \frac{1}{v_2} = \frac{1}{20} + \frac{1}{40} \]

Finding a common denominator (which is 40):

\[ \frac{1}{v_2} = \frac{2}{40} + \frac{1}{40} = \frac{3}{40} \]

Thus,

\[ v_2 = \frac{40}{3} \approx 13.33 \, \text{cm} \]

### Step 5: Determine the nature of the final image
Since \( v_2 \) is positive, it indicates that the final image formed by the converging lens is real and located on the opposite side of the lens from the object.

### Final Answer
The final image is real and located approximately \( 13.33 \, \text{cm} \) from the converging lens.

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