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For this problem on the topic of mirrors and lenses, we are shown in the figure a periscope which can be used for objects that cannot be seen directly.
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It is mainly used in submarines.
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Now we are to suppose that an object is a distance p1 from the upper mirror and the centres of the two flat mirrors are separated by a distance h.
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We want to know the distance of the final image from the lower mirror.
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We want to know if this final image is real or virtual, upright or inverted, its magnification, and we want to know if the image will appear left, right, reversed.
00:38
Now when an object is in front of a plane mirror, that mirror forms an upright virtual image that is the same size of the object and as far behind the mirror as the object is in front of the mirror.
00:48
Now this statement is true even if the mirror is rotated, as shown in the ray diagrams that we have here.
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Now in figure a, which is the figure on the left, the real object o1 is a distance p1 in front of the upper mirror in the periscope.
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This mirror forms the virtual image i1 at a distance p1 behind the mirror, as shown in figure b.
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This image serves as the object for the lower mirror in the periscope and is distance p2 is equal to p1 plus h in front of the lower mirror.
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The lower mirror then forms the final image i2, an upright version.
01:26
Image located a distance p2 which is p1 plus h behind the mirror...