A double concave thin lens made out of glass `(mu=1.5)` have radii of curvature 500cm. This lens is used to rectify the defect in vision of a person. The far point of the person will be at

Options

A. 5 m
B. 2.5 m
C. 1.25 m
D. 1 m

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Refractive index of the lens, \( \mu = 1.5 \)
- Radii of curvature, \( R_1 = -500 \, \text{cm} \) (since it is a concave surface) and \( R_2 = +500 \, \text{cm} \)

### Step 2: Use the Lensmaker's Formula
The Lensmaker's formula is given by:

\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 3: Substitute the values into the formula
Substituting the values of \( \mu \), \( R_1 \), and \( R_2 \):

\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{-500} - \frac{1}{500} \right) \]

### Step 4: Simplify the equation
Calculating \( \mu - 1 \):

\[ \frac{1}{f} = 0.5 \left( \frac{1}{-500} - \frac{1}{500} \right) \]

Calculating the right-hand side:

\[ \frac{1}{f} = 0.5 \left( -\frac{1}{500} - \frac{1}{500} \right) = 0.5 \left( -\frac{2}{500} \right) = 0.5 \times -\frac{2}{500} = -\frac{1}{500} \]

### Step 5: Solve for the focal length \( f \)
Taking the reciprocal to find \( f \):

\[ f = -500 \, \text{cm} \]

### Step 6: Determine the far point of the person
Since the far point of the person is equal to the focal length of the lens used to correct their vision:

\[ \text{Far point} = f = -500 \, \text{cm} \]

### Final Answer
The far point of the person will be at \( 500 \, \text{cm} \) in front of the lens.

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