Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

Virtual, upright, height= $0.5\text{ cm}$

Real, inverted, height= $4\text{ cm}$

Real, inverted, height= $1\text{ cm}$

Virtual, upright, height= $1\text{ cm}$

Step-by-Step Solution

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

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