The magnitudes of the radii of curvature are 30.5 cm and 44.2 cm for the two fac

Question

The magnitudes of the radii of curvature are 30.5 cm and 44.2 cm for the two faces of a biconcave lens. The glass has index of refraction 1.53 for violet light and 1.51 for red light. (a) For a very distant object, locate the image formed by violet light. (Give your answer to two decimal places. Indicate the position of the image relative to the lens with the sign of your answer.) 492 cm (b) For a very distant object, locate the image formed by red light. (Give your answer to two decimal places.Indicate the position of the image relative to the lens with the sign of your answer.) 492 cm

Explanation / Answer

The focal length of the lens is given by

1/f =(n-1)(1/R1 - 1/R2)

= (1.53 -1)(1/-30.5 - 1/44.2)

F = 33.48cm

R1 is negative because the center of curvature of the first surface is on the virtual image side

For a very distant object, p= ? Therefore, q=f = 33.48 cm

The image is virtual, upright and diminshed.

b) The same ray diagram and image characteristics apply for red light.
Again, q = f

1/f =(n-1)(1/R1 - 1/R2)

= (1.51 -1)(1/-30.5 - 1/44.2)

f = -35.39cm

The fact that different wavelengths of light refracted by a lens focus at different points gives rise to
chromatic aberrations

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