7. A 4.60 cm tall object is 28.5 cm in front of a convex mirror with a radius of

Question

7. A 4.60 cm tall object is 28.5 cm in front of a convex mirror with a radius of curvature of 82.0 cm.

a) Where does the image appear to be formed? Is it real or virtual? Upright or inverted? What is the size of the image?

b) A lens is placed 35.0 cm in front of the mirror. It projects an in focus image of the image formed on the mirror on a screen. It is the same height as the original object, but inverted. What is the focal length of the lens? (Note: The image of the object itself formed by the lens is not in focus.)

7. A 4.60 cm tall object is 28.5 cm in front of a convex mirror with a radius of curvature of 82.0 cm Where does the image appear to be formed? Is it real or virtual? Upright or inverted? What is the size of the image? A lens is placed 35.0 cm in front of the mirror. It projects an in focus image of the image formed on the mirror on a screen. It is the same height as the original object, but inverted. What is the focal length of the lens? (Note: The image of the object itself formed by the lens is not in focus.) a) b) 28.5 (b) only cm 5.0 cm..."

Explanation / Answer

a] Use the mirror equation

1/f = 1/v + 1/u

f = R/2 = -82/2 = -41 cm

-1/41 = 1/v + 1/28.5

=> v = - 16.813 cm

the negative sign shows that the image will be virtual and upright.

the size of the image will be:

H = (v/u)h = (16.813/28.5)4.6 = 2.714 cm

b] The object distance will now be: u = 16.813 + 35 = 51.813 cm

Now the final height of the image is 4.6 cm

therefore, the magnification due to the addition of lens will be: M = 4.6/2.714 = 1.695

but M = v/u

=> v = 1.695 x 51.813 = 87.823 cm

now use the lens equation

1/f = 1/v + 1/u

=> 1/f = 1/87.823 + 1/51.813

=> f = 32.587 cm.

This is the focal length of the lens.

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