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Physics · 2025 · 5 marks
CBSE 2025 · Region 7 · Set 1 · Q33
(a) (i) Explain with the help of a labelled ray diagram the formation of final image by an astronomical telescope at infinity. Write the expression…
(i)
Explain with the help of a labelled ray diagram the formation of final image by an astronomical telescope at infinity. Write the expression for its magnifying power.
(ii)
The total magnification produced by a compound microscope is $\displaystyle 20$ . The magnification produced by the eyepiece is $\displaystyle 5$ . When the microscope is focussed on a certain object, the distance between the objective and eyepiece is observed to be $\displaystyle 14$ cm . Calculate the focal lengths of the objective and the eyepiece. (Given that the least distance of distinct vision $\displaystyle =25 \mathrm{~cm}$ )
(i)
Two coherent light waves, each of intensity $\displaystyle \mathrm{I}_{0}$ superpose each other and produce interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is $\displaystyle \phi$. Write its maximum and minimum possible values.
(ii)
In a single slit diffraction experiment, the aperture of the slit is $\displaystyle 3$ mm and the separation between the slit and the screen is $\displaystyle 1.5$ m . A monochromatic light of wavelength $\displaystyle 600$ nm is normally incident on the slit. Calculate the distance of (I) first order minimum, and (II) second order maximum, from the centre of the screen.
Marking-scheme solution
(i)
Light from distant object enters the objective lens & forms a real image A'B' at $\displaystyle f_o$.
This image A'B' acts as an object for eye piece and eye piece forms a magnified image at infinity.
Magnifying Power $\displaystyle = \dfrac{f_o}{f_e}$
(ii)
Image is formed at least distance of distinct vision
$\displaystyle 20 = m_o \times m_e$
$\displaystyle m_o = \dfrac{20}{5} = 4$
$\displaystyle m_e = 1 + \dfrac{D}{f_e}$
$\displaystyle f_e = \dfrac{25}{4}\,cm$
$\displaystyle \dfrac{1}{v_e} - \dfrac{1}{u_e} = \dfrac{1}{f_e}$
$\displaystyle \dfrac{1}{-25} - \dfrac{1}{u_e} = \dfrac{4}{25}$
$\displaystyle u_e = -5\,cm$
$\displaystyle L = v_0 + |u_e|$
$\displaystyle v_0 = 9\,cm$
Given, $\displaystyle \dfrac{v_0}{u_0} = 4$
$\displaystyle \dfrac{1}{v_0} - \dfrac{1}{u_0} = \dfrac{1}{f_0}$
$\displaystyle \dfrac{1}{f_0} = \dfrac{1}{9} - \left(-\dfrac{4}{9}\right)$
$\displaystyle f_0 = \dfrac{9}{5}\,cm$
(i)
$\displaystyle y_1 = a\cos\omega t$
$\displaystyle y_2 = a\cos(\omega t + \phi)$
According to Principle of Superposition
$\displaystyle y = y_1 + y_2$
$\displaystyle = a\left[\cos\omega t + \cos(\omega t + \phi)\right]$
$\displaystyle = 2a\cos\dfrac{\phi}{2}\cos\left(\omega t + \dfrac{\phi}{2}\right)$
$\displaystyle y = A\cos\left(\omega t + \dfrac{\phi}{2}\right)$
where, $\displaystyle A = 2a\cos\dfrac{\phi}{2}$
$\displaystyle I = kA^2$
$\displaystyle I = k\left(4a^2\cos^2\dfrac{\phi}{2}\right)$
$\displaystyle I = 4I_0\cos^2\dfrac{\phi}{2}$
Alternatively: $\displaystyle I = I_1 + I_1 + 2\sqrt{I_1 I_1}\,\cos\phi$
Maximum value $\displaystyle I = 4I_0$
Minimum value $\displaystyle I = 0$
(ii)
(I)
Position of first order minimum
$\displaystyle y = \dfrac{n\lambda D}{a}$
$\displaystyle y_1 = \dfrac{\lambda D}{a}$
$\displaystyle = \dfrac{600\times 10^{-9}\times 1.5}{3\times 10^{-3}} = 3\times 10^{-4}\ \text{m}$
(II)
Position of second order maximum
$\displaystyle y_n = (2n+1)\dfrac{\lambda D}{2a}$
$\displaystyle n = 2,\quad y_2 = \dfrac{5\lambda D}{2a}$
$\displaystyle = \dfrac{5\times 600\times 10^{-9}\times 1.5}{2\times 3\times 10^{-3}} = 7.5\times 10^{-4}\ \text{m}$
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