An object 5.0 cm tall is 14cm cm from a concave lens. The resulting virtual imag

Question

An object 5.0 cm tall is 14cm cm from a concave lens. The resulting virtual image is one-third as large as the object.

A) What is the focal length of the lens? f=______

B) What is the image distance? di=_______

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C) What are the focal length of the lens? f=___________

D) What are the lateral magnification of the image? M=_______

An object 5.0 cm tall is 14cm cm from a concave lens. The resulting virtual image is one-third as large as the object. A) What is the focal length of the lens? f=______ B) What is the image distance? di=_______ ------------------ When an object is placed at 1.9m cm in front of the lens. C) What are the focal length of the lens? f=___________ D) What are the lateral magnification of the image? M=_______ m in front of a diverging lens, a virtual image is formed at 36cm

Explanation / Answer

Hi,

m = v / u = 1/5 ie.

, v = u / 5= 10/5 = 2 cm

then 1/f = 1/u - 1/v, where 'f' is the focal length of the lens

so that 1/f = 1/10 - 1/2

= - 4 / 10 or f

= - 2.5 cm

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