The image of an object placed in front of the collecting lens at a distance of 15 cm is at a distance of 30 cm
The image of an object placed in front of the collecting lens at a distance of 15 cm is at a distance of 30 cm from it. Find the focal length and optical power of the lens. What is the nature of the image in the lens?
To determine the focal length F and the optical power of the lens D, we use the thin lens formula:
1 / F = 1 / d + 1 / f or F = d · f / (d + f), and the definition of “lens power” D = 1 / F.
From the condition of the problem it is known that the image of an object placed in front of the collecting lens at a distance of d = 15 cm = 0.15 m is from it at a distance of f = ± 30 cm = ± 0.3 m. The plus sign is taken for the case when the object and its actual image lie on opposite sides of the lens. The minus sign is taken for the case when the object and its virtual image lie on one side of the lens.
Substitute the values of the quantities into the calculation formula:
F = 0.15 m * 0.3 m / (0.15 m + 0.3 m), F = 0.1 m; D = 1 / 0.1 m; D = 10 diopters – for a valid image;
F = 0.15 m * (- 0.3 m) / (0.15 m + (- 0.3 m)), F = 0.3 m; D = 1 / 0.3 m D ≈ 3.3 diopters – for a virtual image.
Answer: a focal length of 0.1 m and an optical power of 10 diopters for a real image in a lens, a focal length of 0.3 m and an optical power of 3.3 diopters for a virtual image in a lens.
