A converging lens has a focal length of 10.5 cm. Object Distance Image Distance

Question

A converging lens has a focal length of 10.5 cm.

Object Distance Image Distance Real/Virtual Upright/Inverted Enlarged/Same Size/Diminished in Size 4.0 cm cm ---Select---realvirtual ---Select---uprightinverted ---Select---enlargedsame sizediminished in size 14.5 cm cm ---Select---realvirtual ---Select---uprightinverted ---Select---enlargedsame sizediminished in size 17.5 cm cm ---Select---realvirtual ---Select---uprightinverted ---Select---enlargedsame sizediminished in size 20.5 cm cm ---Select---realvirtual ---Select---uprightinverted ---Select---enlargedsame sizediminished in size

Explanation / Answer

Part A)

Apply 1/f = 1/p + 1/q

At 4 cm

1/10.4 = 1/4 + 1/q

q = -6.5 cm

Since q is negative, and larger in magnitude than p (4 cm) we can say the following

The image is virtual, upright, & enlarged

At 14.5 cm

1/10.4 = 1/14.5 + 1/q

q = 36.8 cm

Since q is positive, and larger in magnitude than p (14.5 cm) we can say the following

The image is real, inverted, & enlarged

At 17.5 cm

1/10.4 = 1/17.5 + 1/q

q = 25.6 cm

Since q is positive, and larger in magnitude than p (25.6 cm) we can say the following

The image is real, inverted, & enlarged

At 20.5 cm

1/10.4 = 1/20.5 + 1/q

q = 21.1 cm

Since q is positive, and larger in magnitude than p (20.5 cm) we can say the following

The image is real, inverted, & enlarged

Part B)
At 4 cm

h'/h = -q/p

h'/2 = -(-6.5)/(4)

h' = 3.25 cm

At 20.5 cm

h'/2 = -21.1/20.5

h' = -2.06 cm (The negative just means its inverted, ignore it if you need to, its just sign convention)

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