The far point of a myopic eye is 10 cm from the eye. The focal length of a lens for reading at normal distance ( 25cm) is

Options

A. `-8.55cm`
B. `-16.7cm`
C. `-35.4cm`
D. `-32.7cm`

Explanation

To find the focal length of the lens needed for a myopic eye with a far point of 10 cm, we can follow these steps:

### Step 1: Understand the given information
- The far point of the myopic eye is 10 cm. This means the eye can see objects clearly only up to this distance.
- The normal reading distance (where we want to see clearly) is 25 cm.

### 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 of the lens,
- \( v \) is the image distance,
- \( u \) is the object distance.

### Step 3: Assign values to \( u \) and \( v \)
- Since the object is at a distance of 25 cm from the lens, we have:
\[ u = -25 \, \text{cm} \quad (\text{object distance is negative in lens formula}) \]
- The image should be formed at the far point of the eye, which is at 10 cm. Therefore:
\[ v = -10 \, \text{cm} \quad (\text{image distance is also negative}) \]

### Step 4: Substitute values into the lens formula
Now, substituting the values of \( u \) and \( v \) into the lens formula:
\[ \frac{1}{f} = \frac{1}{-10} - \frac{1}{-25} \]

### Step 5: Simplify the equation
Calculating the right-hand side:
\[ \frac{1}{f} = -\frac{1}{10} + \frac{1}{25} \]
To combine these fractions, find a common denominator (which is 50):
\[ \frac{1}{f} = -\frac{5}{50} + \frac{2}{50} = -\frac{3}{50} \]

### Step 6: Solve for \( f \)
Taking the reciprocal to find \( f \):
\[ f = -\frac{50}{3} \, \text{cm} \]

### Step 7: Convert to decimal form
Calculating the value:
\[ f \approx -16.67 \, \text{cm} \]

### Final Answer
The focal length of the lens required for the myopic eye to read at a normal distance of 25 cm is approximately \( -16.67 \, \text{cm} \).

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