An object is located 26.5 cm from a certain lens. The lens forms a real image th

Question

An object is located 26.5 cm from a certain lens. The lens forms a real image that is twice as high as the object. What is the focal length of this lens? 79.5 cm 8.83 cm 11.3 cm 17.7 cm Now replace the lens used in with another lens. The new lens is a diverging lens whose focal points are at the same distance from the lens as the focal points of the first lens. If the object is 5.00 cm high, what is the height of the image formed by the new lens? The object is still located 26.5 cm from the lens. 12.5 cm 2.0 cm 10.0 cm 10.6 cm 7.5 cm 3.3 cm

Explanation / Answer

PART A

Object distance 'u' = -26 cm
Magnification 'm'= -2

m=[v/u] = f /(u+f)
therefore f / (-26.5 + f) = -2

f = 53 -2f

3f=53
f=53/3=17.77 cm

PART B

The second lens which replaces the first is diverging ( CONCAVE)

AS OBJECT REMAINS SAME u= -26.5 cm
focal length f = -53/3 cm

object height O =5.00 cm
I/O =magnification m =f /(u +f)
I = O f /(u +f )
=5*( - 53/3) / [ -53/2 -53/3]= 2 cm
image will be upright,smaller i.e. 2 cm high

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