You are standing at a distance of 3.50m from a large distance of 20cm toyour right. The distance of your image from mirror is-
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Answer:
The triangle formed by the object (the student) and the rays from the student's head and feet going to the pin-hole and the triangle formed by the image and the rays coming from the pin-hole toward the head and feet of the image are similar triangles so the ratios of corresponding sides of these triangles are equal. That is,
di = 0.067 m = 6.7 cm
26.3 The distance from the cornea to the retina of some particular eye is 2.1 cm. Find the effective focal length of this eye for an object located
(a) at infinity,
(b) for an object located 10 m away, and
(c) for an object located 25 cm away.
We can directly apply the Image Equation
The image distance, di, is the distance from the cornea to the retina, 2.1 cm.
For (a), the object distance, do, is infinite,
f = 2.1 cm
For (b), the object distance is do = 10 m
Watch the units! You can not add meters and centimeters; you can not add inverse meters and inverse centimeters. The two terms on the right must have the same units before you can add them!
f = 2.096 cm 2.1 cm
For (c), the object distance is do = 25 cm
f = 1.94 cm
26.5 An object sits 35 cm to the left of a converging lens with focal length of 20 cm. 75 cm to the right is a second converging lens with focal length of 15 cm. Locate and characterize the final image.
A ray diagram is always a good way to start. There is usually more real understanding available in the ray diagram than in "merely" solving the problem numerically. In addition, a good diagram also provides a place to keep all the dimensions of the problem.
For the first lens, we begin with the Image Equation,
or
And we know f1 = 20 cm and do1 = 35 cm,
di1 = 46.7 cm
This image (which is real, inverted, and enlarged) becomes the object for the second lens as we again apply the Image Equation,
Answer:
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