A convex lens of focal length 100 cm and a concave lens of focal length 10 cm are placed coaxialiy at a separation of 90 cm. If a parallel beam of light is incident on convex lens, then after passing through the two lenses the beam

Options

A. Converges
B. Diverges
C. Remains parallel
D. Disappears

Answer

To solve the problem, we will analyze the behavior of light as it passes through the convex lens and then through the concave lens. Here are the steps to arrive at the solution:

### Step 1: Identify the lenses and their properties
- We have a convex lens with a focal length \( f_1 = 100 \, \text{cm} \).
- We have a concave lens with a focal length \( f_2 = -10 \, \text{cm} \) (the focal length of a concave lens is negative).
- The separation between the two lenses is \( d = 90 \, \text{cm} \).

### Step 2: Determine the behavior of light through the convex lens
- A parallel beam of light is incident on the convex lens.
- According to the lens formula, the rays will converge at the focal point of the convex lens, which is located at a distance of \( 100 \, \text{cm} \) from the lens.
- Since the distance from the convex lens to the concave lens is \( 90 \, \text{cm} \), the rays will converge to a point \( 10 \, \text{cm} \) beyond the concave lens (i.e., \( 100 \, \text{cm} - 90 \, \text{cm} = 10 \, \text{cm} \)).

### Step 3: Analyze the position of the focus relative to the concave lens
- The rays converge at a point \( 10 \, \text{cm} \) to the right of the concave lens.
- This point acts as a virtual object for the concave lens.

### Step 4: Use the lens formula for the concave 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, and \( u \) is the object distance.
- Here, the virtual object distance \( u \) is \( -10 \, \text{cm} \) (negative because it is on the same side as the incoming light).
- The focal length \( f \) of the concave lens is \( -10 \, \text{cm} \).

### Step 5: Calculate the image distance \( v \)
- Plugging in the values into the lens formula:
\[ \frac{1}{-10} = \frac{1}{v} - \frac{1}{-10} \]
\[ \frac{1}{-10} + \frac{1}{10} = \frac{1}{v} \]
\[ 0 = \frac{1}{v} \]
- This implies that \( v \) approaches infinity, meaning that the rays emerge parallel after passing through the concave lens.

### Step 6: Conclusion
- After passing through both lenses, the emergent rays are parallel to the principal axis.

Thus, the final answer is that the emergent beam from the concave lens remains parallel to the principal axis.

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