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.