(1.) A 2.0-cm-tall object is 21 cm in front of a converging lens that has a 32 c

Question

(1.) A 2.0-cm-tall object is 21 cm in front of a converging lens that has a 32 cm focal length.

(A.) Calculate the image position.

(B.) Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.

(2.) A 1.0-cm-tall object is 75 cm in front of a converging lens that has a 35 cm focal length.

(A.) Calculate the image position.

(B.) Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.

(3.) A 2.0-cm-tall object is 14 cm in front of a diverging lens that has a -20 cm focal length.

(A.) Calculate the image position.

(B.) Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.

(4.) A 3.5-cm-tall object is 14 cm in front of a concave mirror that has a 25 cm focal length.

(A.) Calculate the image position.

(B.) Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.

Explanation / Answer

(1) use the basic lens equation:

1/p + 1/q = 1/f

p=object distance = +21 cm
q=image distance
f=focal length = + 32 cm

1/21 + 1/q = 1/32

1/q = 1/32-1/21

q=-61.09 cm

the magnification is given by m = -q/p =61.09/21 = 2.9 (upright)

(2) use the basic lens equation:

1/p + 1/q = 1/f

p=object distance = +75 cm
q=image distance
f=focal length = + 35 cm

1/75 + 1/q = 1/35

1/q = 1/35-1/75

q=65.62 cm

the magnification is given by m = -q/p =-65.62/75 = -0.89 (inverted)

(3) use the thin lens equation

1/i + 1/o = 1/f

where object distance = 14 cm and f =-20cm (convention for diverging lenses is that focal lengths are negative)

then you have

1/i = -1/20 - 1/14

i = -8.24 cm

(4)

use the thin lens equation

1/i + 1/o = 1/f

where object distance = 14 cm and f =-25cm (convention for diverging lenses is that focal lengths are negative)

then you have

1/i = -1/25 - 1/14

i = -8.97 cm

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