CBSE 2022 · Region 5 · Set 1 · Q8 · 3 marks
(a)Define SI unit of power of a lens.(ii)A plano convex lens is made of glass of refractive index 1.5. The radius of curvature of the convex surface is $\displaystyle 25$ cm . (ii.i) Calculate the focal length of the lens. (ii.ii) If an object is placed $\displaystyle 50$ cm in front of the lens, find the nature and position of the image formed.A slit of width $\displaystyle 0.6$ mm is illuminated by a beam of light consisting of two wavelengths $\displaystyle 600$ nm and $\displaystyle 480$ nm. The diffraction pattern is observed on a screen $\displaystyle 1.0$ m from the slit. Find :(i)The distance of the second bright fringe from the central maximum pertaining to light of $\displaystyle 600$ nm.(ii)The least distance from the central maximum at which bright fringes due to both the wavelengths coincide.
(a)
Define SI unit of power of a lens.
(ii)
A plano convex lens is made of glass of refractive index 1.5. The radius of curvature of the convex surface is $\displaystyle 25$ cm . (ii.i) Calculate the focal length of the lens. (ii.ii) If an object is placed $\displaystyle 50$ cm in front of the lens, find the nature and position of the image formed.
A slit of width $\displaystyle 0.6$ mm is illuminated by a beam of light consisting of two wavelengths $\displaystyle 600$ nm and $\displaystyle 480$ nm. The diffraction pattern is observed on a screen $\displaystyle 1.0$ m from the slit. Find :
(i)
The distance of the second bright fringe from the central maximum pertaining to light of $\displaystyle 600$ nm.
(ii)
The least distance from the central maximum at which bright fringes due to both the wavelengths coincide.
Marking-scheme solution
(i)
Dioptre is the reciprocal of focal length of lens in metre.
Alternatively:
$$\text { Dioptre }=\frac{1}{\text { focal length }(\mathrm{m})}
$$Alternatively:
One dioptre is the power of a lens of focal length of one metre
$$\begin{aligned}
& \frac{1}{f}=(\mu-$\displaystyle 1$)\left(\frac{1}{R_{$\displaystyle 1$}}-\frac{1}{R_{$\displaystyle 2$}}\right)
& R_{$\displaystyle 1$}=\infty, \quad R_{$\displaystyle 2$}=-$\displaystyle 25$ \mathrm{~cm}, \mu=$\displaystyle 1$ \cdot $\displaystyle 5$
& \frac{1}{f}=(\mu-$\displaystyle 1$)\left(\frac{1}{\infty}+\frac{1}{25}\right)
& \text { or } \frac{1}{f}=$\displaystyle 0$ \cdot $\displaystyle 5$ \times \frac{1}{25} \text { or } f=$\displaystyle 50$ \mathrm{~cm}
\end{aligned}
$$(ii) $\displaystyle \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Thus the image will be real and inverted
(b)
$$\begin{array}{ll}
\text { Calculation of (i) distance of second bright fringe } & $\displaystyle 1$ \frac{1}{2}
\text { (ii) least distance } & $\displaystyle 1$ \frac{1}{2}
\end{array}
$$
$$\begin{aligned}
& \therefore n=$\displaystyle 3.5$
& \mathrm{x}_{\mathrm{n}}=\frac{(2 n+1) \lambda D}{2 d}=\frac{($\displaystyle 2$ \times $\displaystyle 3.5$+$\displaystyle 1$) \times $\displaystyle 600$ \times $\displaystyle 10$^{-$\displaystyle 9$} \times $\displaystyle 1$}{$\displaystyle 2$ \times $\displaystyle 0$ \cdot $\displaystyle 6$ \times $\displaystyle 10$^{-$\displaystyle 3$}}
& =$\displaystyle 4$ \mathrm{~mm}
\end{aligned}
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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.