4 of 20 tem 4 18.3 cm Constants Previous Answer An object is located 27.5 cm fro

Question

4 of 20 tem 4 18.3 cm Constants Previous Answer An object is located 27.5 cm from a certain lens e lens forms a real image that is twice as high as the object. Correct Part B Now replace the lens used in Part A 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 27.5 cm from the lens View Available Hint(s) 0 12.5 cm O 2.0 cm 10.0 cm 11.0 cm O 7.5 cm 3.3 cm

Explanation / Answer

use the lens equation to find the position of image.

1/f = 1/u + 1/v

If image is real , the image distance di > 0.

The real image is twice the object ,

Us the defination of magnification

M = - di /do = - 2do/do = -2

Magnification negative ,implies image is inverted .

di = 2*27.5 = 55cm

now,

1/f = 1/27.5 +1/55

f = 18.45 cm

Diverging lens with focal length f = -18.45 cm

1/-18.45 = 1/27.5 + 1/di

di = - 11.11 cm

find the image height

M = -(-11.11) / 27.5 = - 0.40

Thus the image height is h = m*ho = 0.40 * 5 = 2.02 cm

Therefore , the image is virtual and 2.02 cm height .

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