Chapter 25: Q25CQ (page 928)
Can a case 1image be larger than the object even though its magnification is always negative? Explain.
Short Answer
Yes, in case one, the image will be larger than the object instead of magnificent being negative.
Chapter 25: Q25CQ (page 928)
Can a case 1image be larger than the object even though its magnification is always negative? Explain.
Yes, in case one, the image will be larger than the object instead of magnificent being negative.
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Get started for freeCombine thin lens equations to show that the magnification for a thin lens is determined by its focal length and the object distance and is given by\(m = f/(f - {d_o})\).
A parallel beam of light containing orange (\(610{\rm{ }}nm\)) and violet (\(410{\rm{ }}nm\)) wavelengths goes from fused quartz to water, striking the surface between them at a\({60.0^ \circ }\)incident angle. What is the angle between the two colours in water?
In Example \(25.7\), the magnification of a book held \(7.50cm\) from a \(10.0cm\) focal length lens was found to be \(3.00\). (a) Find the magnification for the book when it is held \(8.50cm\) from the magnifier. (b) Do the same for when it is held \(9.50cm\) from the magnifier. (c) Comment on the trend in m as the object distance increases as in these two calculations.
On the Moon’s surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. What percent correction is needed to account for the delay in time due to the slowing of light in Earth’s atmosphere? Assume the distance to the Moon is precisely, and Earth’s atmosphere (which varies in density with altitude) is equivalent to a layer 30.0 kmthick with a constant index of refraction n=1.000293.
Some telephoto cameras use a mirror rather than a lens. What radius of curvature mirror is needed to replace afocal length telephoto lens?
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