The radii of curvature of the two surfaces of a lens are 20 cm and 30 cm and the refractive index of the material of the lens is 1.5. If the lens is concave -convex, then the focal length of the lens is

Options

A. 24 cm
B. 10 cm
C. 15 cm
D. 120 cm

Answer

To find the focal length of a concave-convex lens with given radii of curvature and refractive index, we can use the lensmaker's formula:

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

Where:
- \( f \) is the focal length of the lens,
- \( \mu \) is the refractive index of the lens material,
- \( R_1 \) is the radius of curvature of the first surface,
- \( R_2 \) is the radius of curvature of the second surface.

### Step 1: Identify the values given in the problem
- The radius of curvature of the first surface \( R_1 = 20 \, \text{cm} \) (for a convex surface, this is positive).
- The radius of curvature of the second surface \( R_2 = -30 \, \text{cm} \) (for a concave surface, this is negative).
- The refractive index \( \mu = 1.5 \).

### Step 2: Substitute the values into the lensmaker's formula
Using the lensmaker's formula, we substitute the values:

\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-30} \right) \]

### Step 3: Simplify the equation
Calculate \( \mu - 1 \):

\[ \mu - 1 = 1.5 - 1 = 0.5 \]

Now calculate \( \frac{1}{20} - \frac{1}{-30} \):

\[ \frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \]

### Step 4: Substitute back into the equation
Now substitute back into the equation:

\[ \frac{1}{f} = 0.5 \cdot \frac{1}{12} = \frac{0.5}{12} = \frac{1}{24} \]

### Step 5: Calculate the focal length \( f \)
Taking the reciprocal gives us:

\[ f = 24 \, \text{cm} \]

### Final Result
The focal length of the lens is \( 24 \, \text{cm} \).

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