A person cannot see objects clearly 40 cm from him. Can you help him by telling the nature and focal length of lens which he must use to make his eye normal?

Answer

To solve the problem, we need to determine the nature and focal length of the lens that the person should use to see clearly. Here’s the step-by-step solution:

### Step 1: Understand the Problem
The person cannot see objects clearly that are closer than 40 cm. This implies that his near point (the closest distance at which he can see clearly) is at 40 cm. To correct this vision, we need to find a lens that will allow him to see objects at a normal near point (which is typically 25 cm for a normal eye).

### Step 2: Identify the Object and Image Distances
In this case:
- The object distance (u) is at infinity (since we want to correct the vision for distant objects).
- The image distance (v) should be at the person's near point, which is -40 cm (the negative sign indicates that the image is formed on the same side as the object).

### Step 3: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values we have:
- \( v = -40 \, \text{cm} \)
- \( u = \infty \)

Now, substituting these values into the lens formula:
\[ \frac{1}{f} = \frac{1}{-40} - \frac{1}{\infty} \]
Since \( \frac{1}{\infty} = 0 \), the equation simplifies to:
\[ \frac{1}{f} = -\frac{1}{40} \]

### Step 4: Calculate the Focal Length
Now, we can find the focal length (f):
\[ f = -40 \, \text{cm} \]

### Step 5: Determine the Nature of the Lens
Since the focal length is negative, the lens is a **concave lens**.

### Step 6: Calculate the Power of the Lens
The power (P) of a lens is given by the formula:
\[ P = \frac{1}{f} \text{ (in meters)} \]
Converting focal length from cm to meters:
\[ f = -40 \, \text{cm} = -0.40 \, \text{m} \]
Now, substituting this into the power formula:
\[ P = \frac{1}{-0.40} = -2.5 \, \text{diopters} \]

### Final Answer
The person should use a **concave lens** with a focal length of **-40 cm** (or -0.40 m) and the power of the lens is **-2.5 diopters**.

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