Solution (a)
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
The refraction at a spherical surface is governed by the formula:
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$Given data (using sign convention):
- Object distance, $u = -R/3$
- Radius of curvature, $R = +R$
- Refractive index of first medium (air), $n_1 = 1$
- Refractive index of second medium, $n_2 = n$
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
Substituting the values into the formula:
$$\frac{n}{v} - \frac{1}{-R/3} = \frac{n - 1}{R}$$ $$\frac{n}{v} + \frac{3}{R} = \frac{n - 1}{R}$$Isolating the term with $v$:
$$\frac{n}{v} = \frac{n - 1}{R} - \frac{3}{R}$$ $$\frac{n}{v} = \frac{n - 1 - 3}{R} = \frac{n - 4}{R}$$ $$v = \frac{nR}{n - 4}$$Step 3 (Final Concluded Answer): [0.5 Mark]
Since $n < 4$, the value of $(n - 4)$ is negative. Therefore, $v$ is negative.
Result: The image is virtual and formed at a distance of $\frac{nR}{4 - n}$ from the pole on the same side as the object.
💡 EL10 Concept Insight: Think of the curved surface as a lens-like boundary; because the light hits the surface before it can converge, and the medium isn't dense enough to bend the rays to a real focus, the light appears to come from a point behind the object, creating a virtual image.
Solution (b)
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
The intensity at any point in an interference pattern is given by:
$$I = I_0 \cos^2\left(\frac{\phi}{2}\right)$$Where $\phi$ is the phase difference. The relationship between phase difference ($\phi$) and path difference ($\Delta x$) is:
$$\phi = \frac{2\pi}{\lambda} \Delta x$$Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
Given path difference $\Delta x = \lambda/3$. Calculate the phase difference:
$$\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$$Now, substitute $\phi$ into the intensity formula:
$$I = I_0 \cos^2\left(\frac{2\pi/3}{2}\right)$$ $$I = I_0 \cos^2\left(\frac{\pi}{3}\right)$$Since $\cos(\pi/3) = 1/2$:
$$I = I_0 \left(\frac{1}{2}\right)^2 = \frac{I_