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Science · 2024 · 5 marks
CBSE 2024 · Region 4 · Set 1 · Q36
(a) The variation of image distance (v) with object distance (u) for a convex lens is given in the following observation table. Analyse it and answer…
The variation of image distance (v) with object distance (u) for a convex lens is given in the following observation table. Analyse it and answer the questions that follow :
| S. No. | Object distance (u) cm | Image distance (v) cm |
| $\displaystyle 1$ | - $\displaystyle 150$ | +$\displaystyle 30$ |
| $\displaystyle 2$ | -$\displaystyle 75$ | +$\displaystyle 37$•$\displaystyle 5$ |
| $\displaystyle 3$ | -$\displaystyle 50$ | + $\displaystyle 50$ |
| $\displaystyle 4$ | - $\displaystyle 37 \cdot 5$ | +$\displaystyle 75$ |
| $\displaystyle 5$ | -$\displaystyle 30$ | + $\displaystyle 150$ |
| $\displaystyle 6$ | -$\displaystyle 15$ | +$\displaystyle 37$•$\displaystyle 5$ |
(i)
Without calculation, find the focal length of the convex lens. Justify your answer.
(ii)
Which observation is not correct ? Why ? Draw ray diagram to find the position of the image formed for this position of the object.
(iii)
Find the approximate value of magnification for $\displaystyle \mathrm{u}=-30 \mathrm{~cm}$.
(i)
Define principal axis of a lens. Draw a ray diagram to show what happens when a ray of light parallel to the principal axis of a concave lens passes through it.
(ii)
The focal length of a concave lens is $\displaystyle 20$ cm. At what distance from the lens should a $\displaystyle 5$ cm tall object be placed so that its image is formed at a distance of $\displaystyle 15$ cm from the lens ? Also calculate the size of the image formed.
Marking-scheme solution
(i) S. No. $\displaystyle 3$, 2f is $\displaystyle 50$ cm. ∴ 2f = $\displaystyle 50$ cm, or \(\displaystyle \mathrm{f}=25 \mathrm{~cm}\). Justification: Object distance(u) and image distance (v) are same so it implies that object is placed at 2F.
(ii) S. No. $\displaystyle 6$, is not correct.
Reason: For \(\displaystyle \mathrm{u}=-15 \mathrm{~cm}\), sign of v must be - ve ( as the image is formed on the same side of the lens as the object)
Magnification : \(\displaystyle m=\frac{v}{u}\)
(ii)
\(\displaystyle \mathrm{f}=-20 \mathrm{~cm} ; \mathrm{h}=5 \mathrm{~cm}\); \(\displaystyle \mathrm{v}=-15 \mathrm{~cm}\)
\(\displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad\) or \(\displaystyle \frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{(-15)}-\frac{1}{(-20)}\)
\[=\frac{-1}{60 \mathrm{~cm}}
\]
or \(\displaystyle u=-60 \mathrm{~cm}\) object is at a distance of $\displaystyle 60$ cm from the lens
Size of the image(magnification): \(\displaystyle \mathrm{m}=\frac{h^{\prime}}{h}=\frac{v}{u}\)
\[h^{\prime}=\frac{v}{u} \times h=\frac{(-15)}{(-60)} \times 5=1 \cdot 25 \mathrm{~cm}
\]
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