Question

A lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = 169.4 cm. The tip of the object arrow is located at (x,y) = (x1, y1) = (-79.7 cm, 3.83 cm). The index of refraction of the lens is n = 1.41. 1) What is $f_{lens}$, the focal length of the lens? If the lens is converging $f_{lens}$ is positive. It the lens is diverging, $f_{lens}$ is negative. 2) What is $y_2$, the y co-ordinate of the image of the tip of the arrow? 3) The lens is a plano-convex lens. What is $R_{lens}$, the radius of curvature of the convex side of the lens? 4) The object arrow is now moved to x = $x_{1,new}$ = -36.3 cm. What is $x_{2,new}$, the new x co-ordinate of the image of the arrow?

          A lens located in the y-z plane at x = 0 forms an image of an
arrow at x = x2 = 169.4 cm. The tip of the object arrow is
located at (x,y) = (x1, y1) = (-79.7 cm, 3.83 cm). The index
of refraction of the lens is n = 1.41.
1) What is $f_{lens}$, the focal length of the lens? If the lens is converging $f_{lens}$ is positive. It the lens is diverging, $f_{lens}$ is
negative.
2) What is $y_2$, the y co-ordinate of the image of the tip of the arrow?
3) The lens is a plano-convex lens. What is $R_{lens}$, the radius of curvature of the convex side of the lens?
4) The object arrow is now moved to x = $x_{1,new}$ = -36.3 cm. What is $x_{2,new}$, the new x co-ordinate of the image of the
arrow?
        
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A lens located in the y-z plane at x = 0 forms an image of an
arrow at x = x2 = 169.4 cm. The tip of the object arrow is
located at (x,y) = (x1, y1) = (-79.7 cm, 3.83 cm). The index
of refraction of the lens is n = 1.41.
1) What is flens, the focal length of the lens? If the lens is converging flens is positive. It the lens is diverging, flens is
negative.
2) What is y2, the y co-ordinate of the image of the tip of the arrow?
3) The lens is a plano-convex lens. What is Rlens, the radius of curvature of the convex side of the lens?
4) The object arrow is now moved to x = x1,new = -36.3 cm. What is x2,new, the new x co-ordinate of the image of the
arrow?

Added by Purificaci-N R.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A lens located in the y-z plane at x = 0 forms an image of an arrow at x = 169.4 cm. The tip of the object arrow is located at (x, y) (x1, Yi) = (-79.7 cm, 3.83 cm). The index of refraction of the lens is n = 1.41. What is the focal length of the lens (flens)? If the lens is converging, flens is positive. If the lens is diverging, flens is negative. 2) What is Y2, the y-coordinate of the image of the tip of the arrow? 3) The lens is a plano-convex lens. What is the radius of curvature of the convex side of the lens (Rlens)? 4) The object arrow is now moved to X = X1,new = 336.3 cm. What is X2,new, the new x-coordinate of the image of the arrow?
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Transcript

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00:01 Hello student, to solve the given question, let us use the lens formula 1 by f equals 1 by v plus 1 by u.
00:09 Also, magnification is given by minus v by u and height of image divided by height of object.
00:17 So using this relation, we can solve the problem.
00:20 Firstly for part 1, focal length of the lens will be calculated using the lens formula, plug in the values 1 by 1 .1.
00:30 169 .4 plus 1 by 79 .7 equals 1 by f.
00:37 From this calculation we get the final answer for part 1 that is the focal length of lens equals 54 .2 centimeters.
00:48 Similarly using the magnification relation for part 2 m will be calculated as minus 169 .4 divided by 79 .7 equals minus 2 .13.
01:04 Further using the height relation, we get the height of image equals minus 2 .13, multiply by height of object 3 .83, which will be calculated as the final answer for part 2 .2.
01:22 Equals minus 8 .16 centimeters...
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