In an experiment, light shines on a convex lens. Some of the light refracts thro

Question

In an experiment, light shines on a convex lens. Some of the light refracts through the lens, forming a real focus F1. Some of the light reflects off the front surface (which acts like a convex mirror), forming a virtual focus F2.

1) If the index of refraction of the lens material increases, F1 will:
(a) decrease
(b) remain the same
(c) increase

2) If the index of refraction of the lens material increases, F2 will
(a) remain the same
(b) decrease
(c) increase

3) In this procedure: the laser box is placed 14.8 cm from the convex lens, and you find the focal length of the lens is 9.14 cm. If the distance from the lens to the laser box is now 23.9 cm, what will the focal length of the lens be?

Explanation / Answer

(1) If the index of refraction of the lens material increases then from snell's the real focus decreases. (2) If the index of refraction of the lens material increases then from snell's the F2 virtual focus increases. (3) The object distance p = 14.8cm and f = 9.14cm now from lens equation    1/f = 1/p + 1/q then the image distance        q = fp/p-f           = (9.14)(14.8) / 14.8 - 9.14           = 23.899 cm when the object distance is p' = 23.9cm and the image distance q' = 23.899 therefore the focal length                  f = (23.9)(23.9) / 2(23.9)                    = 11.95 cm then from snell's the F2 virtual focus increases. (3) The object distance p = 14.8cm and f = 9.14cm now from lens equation    1/f = 1/p + 1/q then the image distance        q = fp/p-f           = (9.14)(14.8) / 14.8 - 9.14           = 23.899 cm when the object distance is p' = 23.9cm and the image distance q' = 23.899 therefore the focal length                  f = (23.9)(23.9) / 2(23.9)                    = 11.95 cm

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