CBSE 2022 · Region 1 · Set 1 · Q8 · 3 marks
A parallel beam of light of wavelength $\displaystyle 600$ nm is incident normally on a slit of width $\displaystyle 0.2$ mm. If the resulting diffraction pattern is observed on a screen $\displaystyle 1$ m away, find the distance of(i)first minimum, and(ii)second maximum, from the central maximum.A thin equiconvex lens of radius of curvature R made of material of refractive index $\displaystyle \mu_{1}$ is kept coaxially, in contact with an equiconcave lens of the same radius of curvature and refractive index $\displaystyle \mu_{2}\left(>\mu_{1}\right)$. Find :(i)the ratio of their powers, and(ii)the power of the combination and its nature.
A parallel beam of light of wavelength $\displaystyle 600$ nm is incident normally on a slit of width $\displaystyle 0.2$ mm. If the resulting diffraction pattern is observed on a screen $\displaystyle 1$ m away, find the distance of
(i)
first minimum, and
(ii)
second maximum, from the central maximum.
A thin equiconvex lens of radius of curvature R made of material of refractive index $\displaystyle \mu_{1}$ is kept coaxially, in contact with an equiconcave lens of the same radius of curvature and refractive index $\displaystyle \mu_{2}\left(>\mu_{1}\right)$. Find :
(i)
the ratio of their powers, and
(ii)
the power of the combination and its nature.
Marking-scheme solution
(i)
$\displaystyle y = \dfrac{\lambda D}{a}$
$\displaystyle = \dfrac{600\times10^{-9}\times1}{0.2\times10^{-3}}$
$\displaystyle = 3\times10^{-3}\ \text{m} = 3\ \text{mm}$
(ii)
$\displaystyle y = \left(n+\dfrac{1}{2}\right)\dfrac{\lambda D}{a}$
$\displaystyle y = \left(2+\dfrac{1}{2}\right)\dfrac{\lambda D}{a}$
$\displaystyle y = \dfrac{5}{2}\dfrac{\lambda D}{a}$
$\displaystyle y = \dfrac{5}{2}\times\dfrac{600\times10^{-9}\times1}{0.2\times10^{-3}}$
$\displaystyle = 7.5\times10^{-3} = 7.5\ \text{mm}$
(i)
From $\displaystyle P = (\mu-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$
$\displaystyle P_1 = P_{convex} = (\mu-1)\left(\dfrac{1}{R_1}-\left(-\dfrac{1}{R_2}\right)\right)$
$\displaystyle = (\mu-1)\left(\dfrac{2}{R}\right)$
$\displaystyle P_2 = P_{concave} = (\mu-1)\left(-\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$
$\displaystyle = -(\mu-1)\left(\dfrac{2}{R}\right)$
$\displaystyle \therefore \dfrac{P_1}{P_2} = \dfrac{(\mu_1-1)}{-(\mu_2-1)} = \dfrac{(\mu_1-1)}{(1-\mu_2)}$
(ii)
$\displaystyle P = P_1 + P_2$
$\displaystyle = (\mu_1-1)\left(\dfrac{2}{R}\right) + \left(-(\mu_2-1)\right)\left(\dfrac{2}{R}\right)$
$\displaystyle P = \dfrac{2(\mu_1-\mu_2)}{R}$
As $\displaystyle \mu_2 > \mu_1$, $\displaystyle P$ is negative
$\displaystyle \therefore$ Nature is diverging
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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.