1. The focal length of a converging lens in air is +60.0cm. The radii of curvatu

Question

1. The focal length of a converging lens in air is +60.0cm. The radii of curvature of the surfaces comprising the lens are 22.0 cm and 50.0 cm, both having their center of curvatures on the same side. What is the index of refraction of the material which the lens is composed of? a) 1.71 b) 1.33 c)1.65 d) 1.47 e) 0.133 f) 1.28 2. For a singular lens and an upright object, a virtual image will be; a) Inverted and on the back side of the lens bInverted and on the front side of the lens c) Upright and on the backside of the lens d) Upright and on the front side of the lens 3. The focal points of a diverging lens are 25.0 cm from the lens. An object is placed 15.0 cm in front of the lens. What is the magnification of the image produced? a) 0.625 d) 1.26 b)-2.50 e) -1.63 c) -0.728 f 1.89 4. An object is initially at rest, 100 cm from a converging lens with a focal length of 30.0 cm. The object accelerates toward the lens at a constant rate of 0.300 cm/s. At what time will the magnification of the image produced be -2.00? a) 14.4 s b) 8.72s c) 19.1 s d) 17.3 s e) 28.6 s f) 23.3 s 5. Two converging lenses having a focal length of 15.0 cm are placed 25.0 cm apart. Where is the final image located, measured from the 1t lens, if the object is placed 50.0 cm in front of it? a) 4.68 cm behind b) 3.57 cm in front )27.2 cm behind e) 20.3 cm behind f) 35.6 cm behind d) 17.2 cm behind

Explanation / Answer

here,

focal length , f = 60 cm

R1 = 22 cm

R2 = 50 cm

let the refractive index of lens material be n

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

1/60 = (n - 1) * ( 1/22 - 1/50)

solving for n

n = 1.65

the refractive index of lens material is c) 1.65

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