A diverging lens has a focal length of -31.0 cm. An object is placed 19.0 cm in

Question

A diverging lens has a focal length of -31.0 cm. An object is placed 19.0 cm in front of this lens. (a) Calculate the image distance.
1      cm

(b) Calculate the magnification.
2     ?
(c) Is the image real or virtual? real virtual      
(d) Is the image upright or inverted? upright inverted      
(e) Is the image enlarged or reduced in size? reduced enlarged   (a) Calculate the image distance.
1      cm

(b) Calculate the magnification.
2     ?
(c) Is the image real or virtual? real virtual      
(d) Is the image upright or inverted? upright inverted      
(e) Is the image enlarged or reduced in size? reduced enlarged   real virtual       upright inverted       reduced enlarged  

Explanation / Answer

Focal length f = -31 cm

Distance of the object u = 19 cm

From the relation ( 1/ u ) + ( 1/ v ) = ( 1/ f )

0.0526 + ( 1/ v ) = -0.03225

Fromm this distance of the image v = -11.78 cm

(b). magnification m= v / u

= 0.62

(c) virtual as "v" is negative

(d) Upright as m positive

(e) reduced in size as m<1

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