A biconvex lens of refractive index 1.5, has a radius of curvature of magnitude $20 ext{ cm}$. Which one of t

Question

A biconvex lens of refractive index 1.5, has a radius of curvature of magnitude $20	ext{ cm}$. Which one of the following options describes best the image formed of an object of height $2	ext{ cm}$ placed $30	ext{ cm}$ from the lens?

Accepted Answer

1.  **Focal Length of the Lens ($f$):** Using the Lens Maker's formula, $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight)$. For a biconvex lens, by sign convention, $R_1 = +20	ext{ cm}$ and $R_2 = -20	ext{ cm}$.
    $\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}ight) = 0.5 	imes \left(\frac{2}{20}ight) = 0.5 	imes 0.1 = 0.05 = \frac{1}{20}$.
    Thus, $f = +20	ext{ cm}$.
2.  **Image Distance ($v$):** Using the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. The object distance $u = -30	ext{ cm}$.
    $\frac{1}{v} - \left(\frac{1}{-30}ight) = \frac{1}{20} \implies \frac{1}{v} + \frac{1}{30} = \frac{1}{20}$
    $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$.
    Thus, $v = +60	ext{ cm}$. The positive sign indicates that the image forms on the opposite side of the lens, meaning it is a **Real** image.
3.  **Magnification ($m$) and Image Height ($h_i$):** Linear magnification is given by $m = \frac{v}{u}$.
    $m = \frac{+60}{-30} = -2$.
    The negative sign of magnification indicates that the image is **inverted**.
    Now, $m = \frac{h_i}{h_o} \implies h_i = m 	imes h_o = -2 	imes 2	ext{ cm} = -4	ext{ cm}$. The magnitude of the height is $4	ext{ cm}$.
4.  **Conclusion:** The image is real, inverted, and has a height of $4	ext{ cm}$.

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