A point object `O` is placed on the principal axis of a convex lens of focal length `f=20cm` at a distance of 40 cm to the left of it. The diameter of the lens is 10. An eye is placed 60 cm to right of the lens and a distance `h` below the principal axis. The maximum value of `h` to see the image is

Options

A. 0
B. 5 cm
C. 2.5 cm
D. 10 cm

Answer

To solve the problem step by step, we will follow the principles of optics and the lens formula.

### Step 1: Understand the Problem
We have a convex lens with a focal length \( f = 20 \, \text{cm} \) and an object \( O \) placed \( 40 \, \text{cm} \) to the left of the lens. We need to find the maximum height \( h \) below the principal axis at which an eye can be placed \( 60 \, \text{cm} \) to the right of the lens to see the image formed by the lens.

### Step 2: Set Up the Lens Formula
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.

### Step 3: Assign Values
Given:
- \( f = 20 \, \text{cm} \) (positive for a convex lens),
- \( u = -40 \, \text{cm} \) (negative as the object is on the left side of the lens).

### Step 4: Calculate the Image Distance \( v \)
Substituting the values into the lens formula:
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-40} \]
This simplifies to:
\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{40} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{20} - \frac{1}{40} \]
Finding a common denominator (40):
\[ \frac{1}{v} = \frac{2}{40} - \frac{1}{40} = \frac{1}{40} \]
Thus, we find:
\[ v = 40 \, \text{cm} \]

### Step 5: Draw the Ray Diagram
To visualize the situation, we draw a ray diagram. The image \( I \) is formed \( 40 \, \text{cm} \) to the right of the lens.

### Step 6: Determine the Geometry
The eye is placed \( 60 \, \text{cm} \) to the right of the lens, which means the distance from the image to the eye is:
\[ 60 \, \text{cm} - 40 \, \text{cm} = 20 \, \text{cm} \]
The height of the object is negligible since it is a point object, but we need to find the maximum height \( h \) below the principal axis.

### Step 7: Use Similar Triangles
Using the property of similar triangles:
- Let \( AB \) be the height of the image,
- Let \( DB \) be the distance from the lens to the eye (60 cm),
- Let \( EC \) be the distance from the lens to the image (40 cm).

From the similar triangles:
\[ \frac{AB}{DB} = \frac{h}{EC} \]
Substituting the known values:
\[ \frac{AB}{60} = \frac{h}{40} \]
Rearranging gives:
\[ h = \frac{AB \cdot 40}{60} \]

### Step 8: Find the Maximum Height \( h \)
The maximum height \( h \) can be calculated as follows:
Assuming the height of the image \( AB \) is 5 cm (as a reasonable assumption based on the lens diameter):
\[ h = \frac{5 \cdot 40}{60} = \frac{200}{60} = \frac{10}{3} \approx 3.33 \, \text{cm} \]

### Step 9: Conclusion
The maximum value of \( h \) below the principal axis to see the image is approximately \( 3.33 \, \text{cm} \).