Question

A biconvex lens is formed by using two plano-convex lenses as shown in the figure. The refractive index and radius of curvature of surfaces are also mentioned. When an object is placed on the left side of the lens at a distance of \(30\,\text{cm}\), the magnification of the image will be:

• A. \(-2.5\)
• B. \(+2.5\)
• C. \(+2\)
• D. \(-2\)

Hint:

Negative magnification indicates a real and inverted image.

Answer 1

The Correct Option is A

To find the magnification of the image formed by the biconvex lens, we need to calculate the effective focal length of the combined lens system using the formula for the combination of lenses.

Step 1: Calculate the Focal Length of Each Lens

Using the lens maker's formula:

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

where \(\mu\) is the refractive index, and \(R_1, R_2\) are the radii of curvature.

Lens 1:

\(\mu_1 = 1.5, \, R_1 = 15\,\text{cm}, \, R_2 = \infty\)

Substituting the values, we get:

\(\frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{15} - 0 \right)\)

\(f_1 = 30\,\text{cm}\)

Lens 2:

\(\mu_2 = 1.2, \, R_1 = \infty, \, R_2 = -12\,\text{cm}\)

Substituting the values, we get:

\(\frac{1}{f_2} = (1.2 - 1) \left(0 - \frac{1}{-12} \right)\)

\(f_2 = 60\,\text{cm}\)

Step 2: Calculate the Effective Focal Length

The formula for the effective focal length \(f\) of two lenses in contact is:

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(\frac{1}{f} = \frac{1}{30} + \frac{1}{60}\)

\(\frac{1}{f} = \frac{2 + 1}{60} = \frac{3}{60} = \frac{1}{20}\)

\(f = 20\,\text{cm}\)

Step 3: Calculate the Magnification

Using the magnification formula for lenses:

\(M = -\frac{v}{u}\)

Using the lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)

\(\frac{1}{20} = \frac{1}{v} - \frac{1}{-30}\)

\(\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}\)

\(v = 60\,\text{cm}\)

Substituting for magnification \(M\):

\(M = -\frac{60}{-30} = -2\)

Therefore, the correct option is \(-2\).