A diverging lens has a focal length of -24.0 cm. Locate the images for each of t

Question

A diverging lens has a focal length of -24.0 cm. Locate the images for each of the following object distances. For each case, state whether the image is real or virtual and upright or inverted, and find the magnification.

(a) 48.0 cm
cm  --Location of image-- in front of the lens behind the lens no image formed

real, erectreal, inverted     virtual, erectvirtual, inverted

magnification


(b) 24.0 cm
cm  --Location of image-- in front of the lens behind the lens no image formed

real, erectreal, inverted     virtual, erectvirtual, inverted

magnification


(c) 12.0 cm
cm  --Location of image-- in front of the lens behind the lens no image formed

real, erectreal, inverted     virtual, erectvirtual, inverted

magnification

Explanation / Answer

(A) f = -24 cm

p = 48 cm

1/f = 1/p + 1/q

1/(-24) = 1/48 + 1/q

q = - 16 cm

Ans: 16 cm in front of lens

virtual

erect

m = - (-16/48) = 0.333

(B) p = 24 cm

1/-24 = 1/24 + 1/q

q = - 12cm

12cm front of lens

virtual, erect

m = - (-12)/24 = 0.5

(C) p = 12 cm

1/(-24) = 1/12 + 1/q

q = - 8 cm

8 cm in fornt of lens

Virtual, erect

m = - (-8)/12 = 0.67

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