object location, angular magnification & final image

Question

A double-convex lens of 6.25 cm focal length is used as a simple microscope. If it is held close to the eye to form an image 25 cm from the eye, where should the object be located? What is the angular magnification? What would the angular magnification be if the final image were at infinity?

Answer 1

A lens with a focal length (f) of 6.25 cm is held close to the eye, producing a virtual image 25 cm from the eye (at the near point). Assuming negligible lens-eye separation, the image distance (dᵢ) is -25 cm.

1. To determine the object position, we use the thin-lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$. Given $f = 6.25$ cm and $d_i = -25$ cm, we have $\frac{1}{d_o} = \frac{1}{6.25} - \frac{1}{-25} = 0.16 + 0.04 = 0.20$ cm$^{-1}$. Therefore, $d_o = \frac{1}{0.20} = 5.00$ cm. The object should be 5.00 cm in front of the lens.

1. Angular magnification (image at 25 cm, near point) For a simple microscope giving a virtual image at the near point, the angular magnification is

M = 1 + (25 cm / f). Given f = 6.25 cm, M = 1 + (25 / 6.25) = 1 + 4 = 5. Therefore, the angular magnification is 5.

1. Angular magnification at infinity (M∞) is calculated as 25 cm / f. Given M∞ = 6.25 / 25 cm, then M∞ = 4. Answer: 4.