(a)
The image position of the fishes that are located at 5.00 cm
and 25.0 cm
in front of the aquarium wall.
(a)
Answer to Problem 36.36P
Explanation of Solution
Given info: The radius of curvature of the curved plastic is
The formula to calculate image position of the fish inside the aquarium is,
Here,
The radius of curvature will be negative the centre of curvature lies on the object side.
For part (i): when the fish is at
Substitute
The image position for the fish at
For part (ii) when the fish is at
From equation (2) the image position is,
The image is at
Conclusion:
Therefore, the image is
(b)
The magnification of the images for part (a)
(b)
Answer to Problem 36.36P
Explanation of Solution
Explanation
Given info: The radius of curvature of the curved plastic is
The formula to calculate the magnification of the image is,
For part (i): when the fish is at
Substitute
Thus when the fish is at
For part (ii): when the fish is at
Substitute
Thus when the fish is at
Conclusion:
Therefore, when the fish is at
(c)
The reason refractive index of the plastic is not required to solve the problem.
(c)
Answer to Problem 36.36P
Explanation of Solution
Explanation
The plastic has uniform thickness and the surface from which the ray is entering and the surface from which is leaving are parallel to each other. The ray might get slightly displaced, but it will not change the direction of its propagation by going through plastic air interface. The only difference will be due to water-air interface.
Conclusion:
Therefore, the ray might get slightly displaced, but it will not change the direction of its propagation by going through plastic air interface. So the refractive index of plastic is not playing any major role in light propagation.
(e)
The image distance of the fish is greater than the fish itself and the magnification
(e)
Answer to Problem 36.36P
Explanation of Solution
Explanation
For the object distance greater than the radius of curvature the image distance will greater than the distance at which fish is itself. If the aquarium were very long the radius of curvature will not increase therefore if the object distance is more than the radius of curvature the image of the fish will be at even farther distance away from the fish itself.
Conclusion:
Therefore, If the fish is present at distance larger than the radius of curvature the image of the fish would be even farther than that of fish itself.
(d)
The magnification of the image when the image of the fish is even farther than the position of fish itself.
(d)
Answer to Problem 36.36P
Explanation of Solution
For the condition
Formula to calculate the image distance from Lens formula
Substitute
For the condition
Take reciprocal of the above question
Formula to calculate the image distance from Lens formula,
Divide by
The condition is when the image distance is greater than the radius of curvature take the magnitude of the equation.
Substitute
The reciprocal of the equation is
Thus the image of the fish will also be at greater distance than that of radius of curvature.
An example for the above case is let the fish is at twice the distance of the magnitude of radius of curvature.
The image of the fish is calculated from the formula from equation (7).
Substitute
Thus the image of the fish is
The formula to calculate the magnification of the image is,
Substitute
Thus the magnification of the fish image is
Conclusion:
Therefore, if the fish is present at distance larger than the radius of curvature the image of the fish would be even farther than that of fish itself.
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Chapter 36 Solutions
PHYSICS:F/SCI.+ENG.,TECH.UPD.-WEBASSIGN
- Why is the following situation impossible? Consider the lensmirror combination shown in Figure P35.55. The lens has a focal length of fL = 0.200 m, and the mirror has a focal length of fM = 0.500 m. The lens and mirror are placed a distance d = 1.30 m apart, and an object is placed at p = 0.300 m from the lens. By moving a screen to various positions to the left of the lens, a student finds two different positions of the screen that produce a sharp image of the object. One of these positions corresponds to light leaving the object and traveling to the left through the lens. The other position corresponds to light traveling to the right from the object, reflecting from the mirror and then passing through the lens. Figure P35.55 Problem 55 and 57.arrow_forwardIn Figure P26.38, a thin converging lens of focal length 14.0 cm forms an image of the square abcd, which is hc = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c, and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P26.38arrow_forwardTwo converging lenses having focal length of f1 = 10.0 cm and f2 = 20.0 cm are placed d = 50.0 cm apart, as shown in Figure P23.44. The final image is to be located between the lenses, at the position x = 31.0 cm indicated. (a) How far to the left of the first lens should the object be positioned? (b) What is the overall magnification of the system? (c) Is the final image uptight or inserted? Figure P23.44arrow_forward
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- In Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed, which is he = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c. and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb, represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P35.30arrow_forwardAn observer to the right of the mirror-lens combination shown in Figure P36.89 (not to scale) sees two real images that are the same size and in the same location. One image is upright, and the other is inverted. Both images are 1.50 times larger than the object. The lens has a focal length of 10.0 cm. The lens and mirror are separated by 40.0 cm. Determine the focal length of the mirror.arrow_forward
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