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1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.
(a) Find the position of the object.


(b) What is the magnification of the mirror?



2.A contact lens is made of plastic with an index of refraction of 1.45. The lens has an outer radius of curvature of +2.04 cm and an inner radius of curvature of +2.48 cm. What is the focal length of the lens?

Respuesta :

ajinems

The position of the object is = -68cm

The magnification of the mirror= 0.3

Calculation of object distance

The image distance = 20.5cm

The focal length= R/2 = 31.5/2= 15.75

The object distance= ?

Using the lens formula,1/f = 1/v-1/u

1/u = 1/v- 1/f

1/u = 1/20.5 - 1/15.75

1/u = 0.0489- 0.0635

1/u = -0.0146

u = -68cm

The magnification of the mirror is image size/object size

= 20.5cm/-68cm

= 0.3

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