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A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object of height 2 cm placed 30 cm from the lens?
A
Real, inverted, height = 1 cm
B
Virtual, upright, height = 1 cm
C
Virtual, upright, height = 0.5 cm
D
Real, inverted, height = 4 cm
Detailed Solution
The refractive index is not given; taking glass with $\mu = 1.5$.
Lens maker's formula: $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)\left(\frac{1}{20} + \frac{1}{20}\right) = 0.5\times\frac{1}{10} = \frac{1}{20}$
f = 20 cm
Lens formula with u = −30 cm: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, so $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$
v = +60 cm; v is positive, so the image is real and on the other side of the lens.
Magnification $m = \frac{v}{u} = \frac{60}{-30} = -2$; the negative sign means the image is inverted.
Image height = $|m|\times h_o = 2\times2 = 4$ cm
So the image is real, inverted and 4 cm high.
Lens maker's formula: $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)\left(\frac{1}{20} + \frac{1}{20}\right) = 0.5\times\frac{1}{10} = \frac{1}{20}$
f = 20 cm
Lens formula with u = −30 cm: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, so $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$
v = +60 cm; v is positive, so the image is real and on the other side of the lens.
Magnification $m = \frac{v}{u} = \frac{60}{-30} = -2$; the negative sign means the image is inverted.
Image height = $|m|\times h_o = 2\times2 = 4$ cm
So the image is real, inverted and 4 cm high.