Solution (a): Astronomical Refracting Telescope

Step 1 (Ray Diagram & Definition): [1 Mark]

In normal adjustment, the final image is formed at infinity. The parallel rays from a distant object form an image at the focus of the objective lens ($f_o$). This image acts as an object for the eyepiece, which is adjusted such that the final image is at infinity.

[Diagram Description: Objective lens $L_1$ (large aperture) and Eyepiece $L_2$ (small aperture) placed coaxially. Rays from infinity enter $L_1$, converge at $f_o$, and emerge parallel from $L_2$. The angle subtended by the object at the objective is $\alpha$ and the angle subtended by the image at the eye is $\beta$.]

Step 2 (Mathematical Derivation): [1.5 Marks]

The magnifying power $M$ is defined as the ratio of the angle subtended by the image at the eye ($\beta$) to the angle subtended by the object at the objective ($\alpha$):

$$M = \frac{\beta}{\alpha}$$

For small angles, $\tan \beta \approx \beta$ and $\tan \alpha \approx \alpha$. From the geometry of the ray diagram:

$$\tan \alpha = \frac{h}{f_o} \implies \alpha \approx \frac{h}{f_o}$$ $$\tan \beta = \frac{h}{f_e} \implies \beta \approx \frac{h}{f_e}$$

Substituting these into the expression for $M$:

$$M = \frac{h/f_e}{h/f_o}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The magnifying power of an astronomical telescope in normal adjustment is:

$$M = -\frac{f_o}{f_e}$$

(The negative sign indicates that the final image is inverted with respect to the object.)

---

OR

Solution (b): Diffraction at a Single Slit

Step 1 (Intensity Distribution & Governing Law): [1 Mark]

The intensity distribution shows a central maximum of high intensity at the center, flanked by secondary maxima of decreasing intensity and minima where intensity is zero.

Condition for minima: $a \sin \theta = n\lambda$ (where $n = \pm 1, \pm 2, ...$)

Step 2 (Mathematical Working for Maxima): [1.5 Marks]

The secondary maxima occur approximately midway between the minima. The condition for the $n^{th}$ secondary maximum is given by:

$$a \sin \theta = (n + \frac{1}{2})\lambda$$

For the first secondary maximum, we set $n = 1$:

$$a \sin \theta = \frac{3\lambda}{2}$$

Since $\theta$ is very small, $\sin \theta \approx \theta \approx \frac{y}{D}$:

$$a \left( \frac{y}{D} \right) = \frac{3\lambda}{2}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The position of the first secondary maximum from the central maximum is:

$$y = \frac{3\lambda D}{2a}$$

💡 EL10 Concept Insight:

Think of the slit like a row