A diverging lens has a focal length of -22.0 cm. Locate the images for each of t
ID: 1519209 • Letter: A
Question
A diverging lens has a focal length of -22.0 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) 44.0 cm cm real, erect real, inverted virtual, upright virtual, inverted magnification cross product
(b) 22.0 cm cm real, erect real, inverted virtual, upright virtual, inverted magnification cross product
(c) 11.0 cm cm real, erect real, inverted virtual, upright virtual, inverted magnification cross product
Explanation / Answer
here,
the focal length of lens , f = - 22 cm
(a)
when the object distance ,do = 44 cm
let the image distance be di
using the lens formula
1/f = 1/do + 1/di
- 1/22 = 1/44 + 1/di
di = - 14.66 cm
then the image is upright virtual
(b)
when the object distance ,do = 22 cm
let the image distance be di
using the lens formula
1/f = 1/do + 1/di
- 1/22 = 1/22 + 1/di
di = - 11 cm
then the image is upright virtual
(c)
when the object distance ,do = 11 cm
let the image distance be di
using the lens formula
1/f = 1/do + 1/di
- 1/22 = 1/11 + 1/di
di = - 7.33 cm
then the image is upright virtual
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