Point object O is placed on the principal axis of a convex lens of focal length 20cm at a distance of 40 cm to the left of it. The diameter of the lens is 10cm to the right of the lens at a distance h below the principal axis, then the maximum value of h to see the image will be

Options

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

Explanation

To solve the problem step by step, we will use the lens formula and the concept of similar triangles.

### Step 1: Identify the given data
- Focal length of the convex lens (f) = 20 cm
- Distance of the object (u) = -40 cm (the negative sign indicates that the object is on the same side as the incoming light)
- Diameter of the lens = 10 cm

### Step 2: Use the lens formula to find the image distance (v)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-40} \]
This simplifies to:
\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{40} \]
To combine the fractions on the right, we need a common denominator:
\[ \frac{1}{20} = \frac{2}{80} + \frac{1}{80} = \frac{3}{80} \]
Now, equating both sides:
\[ \frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40} = \frac{1}{40} \]
Thus,
\[ v = 40 \text{ cm} \]

### Step 3: Determine the height of the image (h')
Using the magnification formula:
\[ \text{Magnification} (m) = \frac{h'}{h} = -\frac{v}{u} \]
Where \(h'\) is the height of the image and \(h\) is the height of the object (which we will consider as 5 cm for calculation purposes). Thus:
\[ m = -\frac{40}{-40} = 1 \]
This means that the height of the image is equal to the height of the object:
\[ h' = 1 \times h = 5 \text{ cm} \]

### Step 4: Determine the maximum height (h) below the principal axis
Since the image is formed at a distance of 40 cm to the right of the lens, and the diameter of the lens is 10 cm, the maximum height \(h\) below the principal axis can be calculated using similar triangles.

Using the property of similar triangles:
\[ \frac{H}{20} = \frac{5}{40} \]
Cross-multiplying gives:
\[ H = \frac{5 \times 20}{40} = 2.5 \text{ cm} \]

### Final Answer
The maximum value of \(h\) below the principal axis to see the image is **2.5 cm**.

---