Concept explainers
Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F1 and F2 of the lens are 5.00 cm from the center of the lens, (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the filial image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.
(a)
The index of refraction of the lens material.
Answer to Problem 36.95CP
The index of refraction of the lens material is
Explanation of Solution
Given data: The radii of curvature for the converging Len’s surfaces are
Write the expression for the focal length of the lens,
Here,
Substitute
Conclusion:
Therefore the index of refraction of the lens material is
(b)
The position of the final image.
Answer to Problem 36.95CP
The position of the final image is
Explanation of Solution
Given data: The radii of curvature for the converging Len’s surfaces are
The expression for the focal length is,
Here,
Substitute
Thus the focal length is
The expression for the lens equation is,
Here,
Substitute
Thus the image distance from the lens is
The expression for the object distance for mirror is,
Here,
Substitute
Thus the object distance for mirror is
The expression for the mirror equation is,
Here,
Substitute
Thus the image distance from the mirror is
The expression for the object distance for lens is,
Substitute
Thus the object distance for lens is
The expression for the lens equation is,
Here,
Substitute
Conclusion:
Therefore the position of the final image is
(c)
The magnification as seen by the eye in the figure.
Answer to Problem 36.95CP
The magnification as seen by the eye in the figure is
Explanation of Solution
Given data: The radii of curvature for the converging Len’s surfaces are
The overall magnification factor is,
Substitute
Conclusion:
Therefore the magnification as seen by the eye in the figure is
(d)
Whether the final image is inverted or upright.
Answer to Problem 36.95CP
The final image is inverted.
Explanation of Solution
Given data: The radii of curvature for the converging Len’s surfaces are
From part (c) of the question the value of the magnification factor is,
It is known that the image will be upright when the magnification factor is positive and the image will be inverted when the magnification factor is negative.
Since the magnification factor is negative so the image will be upright.
Conclusion:
Therefore the final image is inverted.
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Chapter 36 Solutions
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