Question

Three plano-convex lenses A, B, and C are combined in pairs to form Newton's ring interference apparatus. A monochromatic light with a wavelength of $\lambda$ = 600nm is incident vertically. Observing the dark rings caused by the air layer between the lenses, when lenses A and B are combined as shown in figure (i), the radius of the $10^{th}$ dark ring is $r_{AB}$ = 4.0 mm. When lenses B and C are combined as shown in figure (ii), the radius of the $10^{th}$ dark ring is $r_{BC}$ = 4.5 mm and when lenses C and A are combined as shown in figure (iii), the radius of the $10^{th}$ dark ring is $r_{CA}$ = 5.0 mm.

• A. The radius of curvature of lens A is approximately 6.28 m.
• B. The radius of curvature of lens B is approximately 4.64 m.
• C. The radius of curvature of lens C is approximately 6.2 m.
• D. The radius of curvature of lens C is approximately 3.1 m.

Answer 1

Answer: (A), (B)

a, b

Answer 2

In a Newton's ring experiment formed by combining two plano-convex lenses with radii of curvature $R_1$ and $R_2$ and placing their curved surfaces in contact, the effective radius of curvature of the air film is given by $R_{eff} = \frac{R_1 R_2}{R_1 + R_2}$, which can also be written as $\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}$.

The radius of the $n^{th}$ dark ring is given by $r_n^2 = n \lambda R_{eff}$.

Let $R_A$, $R_B$, and $R_C$ be the radii of curvature of lenses A, B, and C, respectively. The wavelength of light is $\lambda = 600$ nm $= 600 \times 10^{-9}$ m. The order of the dark ring is $n=10$.

Case 1: Lenses A and B combined. Radius of 10th dark ring $r_{AB} = 4.0$ mm $= 4.0 \times 10^{-3}$ m. $r_{AB}^2 = 10 \lambda R_{eff, AB}$, where $\frac{1}{R_{eff, AB}} = \frac{1}{R_A} + \frac{1}{R_B}$. $(4.0 \times 10^{-3})^2 = 10 \times (600 \times 10^{-9}) \times R_{eff, AB}$ $16 \times 10^{-6} = 6000 \times 10^{-9} \times R_{eff, AB} = 6 \times 10^{-6} \times R_{eff, AB}$ $R_{eff, AB} = \frac{16 \times 10^{-6}}{6 \times 10^{-6}} = \frac{16}{6} = \frac{8}{3}$ m. So, $\frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{R_{eff, AB}} = \frac{3}{8}$.

Case 2: Lenses B and C combined. Radius of 10th dark ring $r_{BC} = 4.5$ mm $= 4.5 \times 10^{-3}$ m. $r_{BC}^2 = 10 \lambda R_{eff, BC}$, where $\frac{1}{R_{eff, BC}} = \frac{1}{R_B} + \frac{1}{R_C}$. $(4.5 \times 10^{-3})^2 = 10 \times (600 \times 10^{-9}) \times R_{eff, BC}$ $20.25 \times 10^{-6} = 6 \times 10^{-6} \times R_{eff, BC}$ $R_{eff, BC} = \frac{20.25 \times 10^{-6}}{6 \times 10^{-6}} = \frac{20.25}{6} = 3.375 = \frac{27}{8}$ m. So, $\frac{1}{R_B} + \frac{1}{R_C} = \frac{1}{R_{eff, BC}} = \frac{8}{27}$.

Case 3: Lenses C and A combined. Radius of 10th dark ring $r_{CA} = 5.0$ mm $= 5.0 \times 10^{-3}$ m. $r_{CA}^2 = 10 \lambda R_{eff, CA}$, where $\frac{1}{R_{eff, CA}} = \frac{1}{R_C} + \frac{1}{R_A}$. $(5.0 \times 10^{-3})^2 = 10 \times (600 \times 10^{-9}) \times R_{eff, CA}$ $25 \times 10^{-6} = 6 \times 10^{-6} \times R_{eff, CA}$ $R_{eff, CA} = \frac{25 \times 10^{-6}}{6 \times 10^{-6}} = \frac{25}{6}$ m. So, $\frac{1}{R_C} + \frac{1}{R_A} = \frac{1}{R_{eff, CA}} = \frac{6}{25}$.

We have a system of linear equations for $\frac{1}{R_A}$, $\frac{1}{R_B}$, and $\frac{1}{R_C}$:

1. $\frac{1}{R_A} + \frac{1}{R_B} = \frac{3}{8}$
2. $\frac{1}{R_B} + \frac{1}{R_C} = \frac{8}{27}$
3. $\frac{1}{R_C} + \frac{1}{R_A} = \frac{6}{25}$

Adding the three equations: $2 \left(\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C}\right) = \frac{3}{8} + \frac{8}{27} + \frac{6}{25} = \frac{3 \times 27 \times 25 + 8 \times 8 \times 25 + 6 \times 8 \times 27}{8 \times 27 \times 25} = \frac{2025 + 1600 + 1296}{5400} = \frac{4921}{5400}$. $\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C} = \frac{4921}{10800}$.

Subtracting equation (2) from this sum: $\frac{1}{R_A} = \left(\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C}\right) - \left(\frac{1}{R_B} + \frac{1}{R_C}\right) = \frac{4921}{10800} - \frac{8}{27} = \frac{4921}{10800} - \frac{8 \times 400}{27 \times 400} = \frac{4921 - 3200}{10800} = \frac{1721}{10800}$. $R_A = \frac{10800}{1721} \approx 6.2754$ m.

Subtracting equation (3) from the sum: $\frac{1}{R_B} = \left(\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C}\right) - \left(\frac{1}{R_C} + \frac{1}{R_A}\right) = \frac{4921}{10800} - \frac{6}{25} = \frac{4921}{10800} - \frac{6 \times 432}{25 \times 432} = \frac{4921 - 2592}{10800} = \frac{2329}{10800}$. $R_B = \frac{10800}{2329} \approx 4.6363$ m.

Subtracting equation (1) from the sum: $\frac{1}{R_C} = \left(\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C}\right) - \left(\frac{1}{R_A} + \frac{1}{R_B}\right) = \frac{4921}{10800} - \frac{3}{8} = \frac{4921}{10800} - \frac{3 \times 1350}{8 \times 1350} = \frac{4921 - 4050}{10800} = \frac{871}{10800}$. $R_C = \frac{10800}{871} \approx 12.40$ m.

Now let's check the options: a. The radius of curvature of lens A is approximately 6.28 m. Our calculated $R_A \approx 6.2754$ m, which is approximately 6.28 m. This option is correct. b. The radius of curvature of lens B is approximately 4.64 m. Our calculated $R_B \approx 4.6363$ m, which is approximately 4.64 m. This option is correct. c. The radius of curvature of lens C is approximately 6.2 m. Our calculated $R_C \approx 12.40$ m. This option is incorrect. d. The radius of curvature of lens C is approximately 3.1 m. Our calculated $R_C \approx 12.40$ m. This option is incorrect.

Both options a and b are correct. Since it is a multiple choice question, and usually in JEE/NEET there is only one correct option, let's re-examine the problem and calculations. However, based on the provided options and our calculations, both a and b are valid answers. Assuming this is a multiple correct options question, both a and b are correct. If it is a single correct option question, there might be a typo in the options or the problem statement. Assuming it is a multiple correct option question.