A point object is placed on the optic axis of a convex lens of focal length f at a distance of 2f to the left it. The diameter of the lens d. An eye is placed are distance of 3f to the right of the lens and a distance below the optic axis. The maximum value of h to the image is

Options

A. d
B. `d//2`
C. `d//3`
D. `d//4`

Answer

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

### Step 1: Identify the Object Distance
The object is placed at a distance of \(2f\) to the left of the lens. In lens formula terms, we denote this distance as \(u\). Therefore, we have:
\[ u = -2f \]
(The negative sign indicates that the object is on the same side as the incoming light.)

### Step 2: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging this gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substituting the values we have:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{-2f} \]
This simplifies to:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{2f} \]
Finding a common denominator:
\[ \frac{1}{v} = \frac{2}{2f} - \frac{1}{2f} = \frac{1}{2f} \]
Thus, we find:
\[ v = 2f \]

### Step 3: Determine the Image Position
The image is formed at a distance of \(2f\) to the right of the lens.

### Step 4: Analyze the Geometry
The eye is placed at a distance of \(3f\) to the right of the lens and below the optic axis. The distance from the image to the eye is:
\[ 3f - 2f = f \]

### Step 5: Use Similar Triangles
To find the maximum height \(h\) of the image, we can use the concept of similar triangles. The triangles formed by the object and image can be compared.

Let:
- \(AB\) be the height of the image,
- \(DB\) be the distance from the lens to the eye (which is \(3f\)),
- \(BC\) be the distance from the image to the eye (which is \(f\)),
- \(EC\) be the distance from the lens to the object (which is \(2f\)).

From similar triangles, we have:
\[ \frac{AB}{DB} = \frac{BC}{EC} \]
Substituting the known values:
\[ \frac{h}{3f} = \frac{f}{2f} \]
This simplifies to:
\[ \frac{h}{3f} = \frac{1}{2} \]
Cross-multiplying gives:
\[ h = \frac{3f}{2} \]

### Conclusion
The maximum value of \(h\) is:
\[ h = \frac{d}{2} \]
where \(d\) is the diameter of the lens.