CBSE 2022 · Region 2 · Set 1 · Q6 · 3 marks
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.
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
$$m = \frac{\tan\beta}{\tan\alpha} \approx \frac{\beta}{\alpha} \quad \text{(as } \alpha, \beta \text{ are small angles)}$$
$$\approx \frac{h}{f_e} \times \frac{f_0}{h}$$
$$= \frac{f_0}{f_e}$$
(b)
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 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.