Chapter 21, 22,23 Light-Relection and Retreet 0 23.P.033. A diverging lens has a

Question

Chapter 21, 22,23 Light-Relection and Retreet 0 23.P.033. A diverging lens has a focal length of magnitode 16.8 (a) Locate the images for each of the following object distances 33.6 cm distance location P- 6.8 cm suu 8.4 cm location Select (b) Is the image for the object at distance 33.6 real or virtual? O real virtual Is the image for the object at distance 16.8 real or virtual? real O virtual Is the image for the object at distance 8.4 real or virtual? O real O virtual (c) Is the image for the object at distance 33.6 upright or inverted? upright O inverted Is the image for the object at distance 16.8 upright or inverted? upright inverted Is the image for the object at distance 8.4 upright or inverted? upright

Explanation / Answer

a)

f=-16.8cm, p=33.6cm

Use eqn,

1/f=1/p+1/q

-1/16.8=1/33.6+1/q

q=-11.2cm

Negative sign indicates image is on the same side as the object is.

Also it indicates image is virtual.

Magnification=m=-q/p = -(-11.2)/33.6 = 0.33

Positive sign of magnification indicates image is upright.

For : f=-16.8cm, p=16.8cm

Use eqn,

1/f=1/p+1/q

-1/16.8=1/16.8+1/q

q=-8.4cm

Negative sign indicates image is on the same side as the object is.

Also it indicates image is virtual.

Magnification=m=-q/p = -(-8.4)/16.8 = 2.0

Positive sign of magnification indicates image is upright.

For : f=-16.8cm, p=8.4cm

Use eqn,

1/f=1/p+1/q

-1/16.8=1/8.4+1/q

q=-5.6cm

Negative sign indicates image is on the same side as the object is.

Also it indicates image is virtual.

Magnification=m=-q/p = -(-5.6)/8.4 = 0.67

Positive sign of magnification indicates image is upright.

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