Concept explainers
(a)
The value of
(a)
Answer to Problem 50P
The value of
Explanation of Solution
Write the expression for the relation between the object distance, image distance and focal length.
Here,
Write the expression for magnification.
Here,
Write the expression for the relation between height and magnification.
Here,
Substitute
Conclusion:
Substitute
Substitute
Substitute
Substitute
Therefore, the value of
(b)
The sketch of the image.
(b)
Answer to Problem 50P
The sketch of the image is given in figure 1.
Explanation of Solution
The following figure gives the diagrammatic representation of the image formed by the thin lens.
Conclusion:
Therefore, the sketch of the image is given in figure 1.
(c)
To demonstrate that the area is
(c)
Answer to Problem 50P
It is demonstrated that the area is
Explanation of Solution
Substitute
Conclusion:
Substitute
Write the expression for the modulus of the above equation.
Therefore, it is demonstrated that the area is
(d)
The reason for the area to be
(d)
Answer to Problem 50P
The integral will add up the area of the whole image.
Explanation of Solution
The vertical dimension of the image is given by
Therefore, the integral will add up the area of the whole image.
(e)
The area of the image.
(e)
Answer to Problem 50P
The area of the image is
Explanation of Solution
Write the expression for the area of the image.
Conclusion:
Substitute
Therefore, the area of the image is
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Chapter 36 Solutions
PHYSICS:F/SCI...W/MOD..-UPD(LL)W/ACCES
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