Question

An equiconvex lens of radius of curvature 14 cm is made up of two different materials. Left half and right half of vertical portion is made up of material of refractive index 1.5 and 1.2 respectively as shown in the figure. If a point object is placed at a distance of 40 cm , calculate the image distance.

• A. 25 cm
• B. 50 cm
• C. 35 cm
• D. 40 cm

Answer 1

The Correct Option is D

The lens maker's formula is given by:

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

Where:

• \( f \) = focal length
• \( n \) = refractive index of the lens material
• \( R_1 \) = radius of curvature of the first surface
• \( R_2 \) = radius of curvature of the second surface

2. Applying the Formula for Each Half

• Left Half (n = 1.5): \( R_1 = +14 \) cm and \( R_2 = -14 \) cm

\( \frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{14} - \frac{1}{-14} \right) = 0.5 \cdot \frac{2}{14} = \frac{1}{14} \)

So, \( f_1 = 14 \) cm

• Right Half (n = 1.2): \( R_1 = +14 \) cm and \( R_2 = -14 \) cm

\( \frac{1}{f_2} = (1.2 - 1) \left( \frac{1}{14} - \frac{1}{-14} \right) = 0.2 \cdot \frac{2}{14} = \frac{1}{35} \)

So, \( f_2 = 35 \) cm

3. Effective Focal Length

Since the lens is split vertically, we can treat each half as a separate lens. The power of the combined lens is the sum of the powers of each half. The power P= 1/f and we know that power of lenses in contact is the sum of their individual powers

\(P = P_1+P_2\)

\(\frac{1}{F} = \frac{1}{2f_1} + \frac{1}{2f_2}\), Note the 1/2 factor because each half is only half the area of a full lens.

\(\frac{1}{F} = (\frac{1}{2})(\frac{1}{14} + \frac{1}{35})\)

\(\frac{1}{F} = (\frac{1}{2})((\frac{5+2}{70})) = \frac{1}{2}(\frac{7}{70}) = \frac{1}{20}\)

\(F = 20\)

4. Using the Lens Formula

Now we use the lens formula to find the image distance (v):

\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

Where:

• \( f \) = effective focal length (20 cm)
• \( v \) = image distance (what we want to find)
• \( u \) = object distance (-40 cm)

\(\frac{1}{20} = \frac{1}{v} - \frac{1}{-40}\)

\(\frac{1}{20} = \frac{1}{v} + \frac{1}{40}\)

\(\frac{1}{v} = \frac{1}{20} - \frac{1}{40}\)

\(\frac{1}{v} = \frac{2-1}{40}\)

\(\frac{1}{v} = \frac{1}{40}\)

\(v = 40\) cm

Answer

The image distance is 40 cm.

Answer 2

Given, the lens is equiconvex with a radius of curvature \( R = 14 \) cm. The lens has two materials with refractive indices 1.5 and 1.2 for the left and right halves, respectively. We will use the lens maker's formula to calculate the focal length of the lens. The lens maker's formula is: \[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] For an equiconvex lens, \( R_1 = R \) (curvature of the left surface) and \( R_2 = -R \) (curvature of the right surface). The formula becomes: \[ \frac{1}{f} = (\mu_1 - 1) \left( \frac{1}{R_1} \right) + (\mu_2 - 1) \left( \frac{1}{R_2} \right) \] Where \( \mu_1 = 1.5 \) for the left half and \( \mu_2 = 1.2 \) for the right half. 1. For the left portion (material with refractive index 1.5): \[ \frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{14} \right) = \frac{0.5}{14} \] 2. For the right portion (material with refractive index 1.2): \[ \frac{1}{f_2} = (1.2 - 1) \left( \frac{-1}{14} \right) = \frac{-0.2}{14} \] So, the total effective focal length is the sum of the two contributions: \[ \frac{1}{f} = \frac{0.5}{14} + \frac{-0.2}{14} = \frac{0.3}{14} \] Thus, \[ f = \frac{14}{0.3} = 46.67 \, \text{cm} \] Now, using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Where \( u = -40 \, \text{cm} \) (object distance). Substituting the values: \[ \frac{1}{46.67} = \frac{1}{v} - \frac{1}{-40} \] \[ \frac{1}{v} = \frac{1}{46.67} + \frac{1}{40} \] \[ \frac{1}{v} = \frac{1}{40} + \frac{1}{46.67} = \frac{46.67 + 40}{40 \times 46.67} = \frac{86.67}{1866.8} = 0.0465 \] Thus, \[ v = \frac{1}{0.0465} \approx 40 \, \text{cm} \] Therefore, the image distance is approximately 40 cm.