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Physics · 2022 · 3 marks
CBSE 2022 · Region 2 · Set 1 · Q6
(a) Draw a labelled ray diagram showing the formation of an image by an astronomical refracting telescope in normal adjustment. Hence, obtain the…
Draw a labelled ray diagram showing the formation of an image by an astronomical refracting telescope in normal adjustment. Hence, obtain the expression for its magnifying power.
A plane wavefront of light of wavelength ' $\displaystyle \lambda$ ' is incident normally on a narrow slit of width 'a' and a diffraction pattern is observed on a screen at a distance 'D' from the slit.
(i)
Depict the intensity distribution in the pattern observed.
(ii)
Obtain the expression for the first maximum from the central maximum.
Marking-scheme solution
$$m = \frac{\tan\beta}{\tan\alpha} \approx \frac{\beta}{\alpha} \quad \text{(as } \alpha, \beta \text{ are small angles)}$$
$$\approx \frac{h}{f_e} \times \frac{f_0}{h}$$
$$= \frac{f_0}{f_e}$$
(i)
Intensity distribution curve
For maximum
$$a\sin\theta = \left(n + \frac{1}{2}\right)\lambda$$
For first Maximum ; n = $\displaystyle 1$
For small $\displaystyle \theta$; $\displaystyle a\theta = \dfrac{3\lambda}{2} \Rightarrow \theta = \dfrac{3\lambda}{2a}$
$$\therefore \frac{x}{D} = \frac{3}{2}\frac{\lambda}{a}$$
$$\therefore x = \frac{3}{2}\frac{\lambda D}{a}$$
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CBSE Class 12 Physics past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.