A diverging lens has a focal length of 10.5 cm . For each of two objects located
ID: 1985110 • Letter: A
Question
A diverging lens has a focal length of 10.5 cm . For each of two objects located to the left of the lens, one at a distance of = 18.0 cm and the other at a distance of = 7.50 cm, determine?Part A
the image position.
Enter your answer as two numbers separated with a comma.
, =cm
Part B
the magnification.
Enter your answer as two numbers separated with a comma.
, = cm
Part C
whether the image is real or virtual.
Both images are real.
The image of the first object is real and the image of the second object virtual.
The image of the first object is virtual and the image of the second object real.
Both images are virtual.
Part D
whether the image is erect or inverted.
Both images are erect.
The image of the first object is erect and the image of the second object inverted.
The image of the first object is inverted and the image of the second object erect.
Both images are inverted.
Explanation / Answer
Given
The lens is a diverging lens so the focal length is f = -10.5cm
here first the object distance is p=18.0cm
we have the thin lens equation as
1/f = 1/p+/1q
1/q = 1/f - 1/p
or q = fp/p-f
= (-10.5cm)(18.0cm)/18.0cm+10.5cm
= -6.6315 cm
for the object distance at p=7.50cm
we have
q = fp/p-f
= (-10.5cm)(7.50cm)/7.50cm+10.50cm
= -4.375cm
The magnification for q= -6.6315 cm is M=-q/p
= -(-6.6315cm)/18.0cm
= 0.36841667
As the magnification is positive the image will be upright and as q is negative the image is virtual.
The magnification for q = -4.375 cm is M=-q/p
= -(-4.375 cm)/7.50 cm
= 0.583333
As the magnification is positive the image will be upright and as q is negative the image is virtual.
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