b-A 1.0-cm-tall object is 60 cm in front of a diverging lens that has a -30 cm focal length.Two converging lenses with focal lengths of 40 cm and 20 cm are 16 cm apart. A diverging lens has a focal length of -30.0 cm. At all times while playing the tape passes over the tape head at a constant speed of 4.75 cm/sec. a. At the beginning of a tape, the radius of one wheel is 2.5cm and...A 1.0-cm-tall object is 60 cm in front of a diverging lens that has a -30 cm focal length. Calculate the image position.
A 1.0-cm-tall object is 13 cm in front of a converging lens that has...
In the given problem, there is a diverging lens whose focal length has given us minus 30 centimeter the height of the object which has been kept in front. This is F one the first principle focus off this concave lens. Now, where we put the object at a distance of 60 centimeter means just double off this.A 2.-cm-tall object is 15 cm in front of a diverging lens that has a -20 cm focal length.a. Use ray tracing to find the position and height of the image. To do this accurately, use a ruler or paper with a grid. Determine the image distance and image height by making measurements on your diagram.b...A 1.0-cm-tall object is 95 cm in front of a converging lens that has a 35 cm focal length.
Cm Tall Object Cm Front Diverging Lens Cm Focal Length
Calculate The Image Position And Height. This problem has been solved!
Video Transcript
hello. In the given downside, there is a diverging lens whose focal duration (*60*) given us minus 30 centimeter the height of the object which (*60*) been stored in front. Off it is 1.Zero centimeter in this object (*60*) been saved at a distance off 60 centimeter in front of this motive force's England's, which is written as minus 60 centimeter. So in the first section of the problem, we have to find the gap off the image and the height off the picture with the assist off changing. So in order to draw the Rada flooring, first of all, they draw the main get right of entry to. Hence, we used our scale to draw this essential axis here. This is a long principal axis. As we've got noticed, we have toe put the object at a distance off 60 centimeter. Quite huge distance. That's why you need to take are reasonably long most important axes at this foremost axes. Suppose this is the purpose. See, representing the optical middle off the concave lens. Now we're going to draw up concave lens right here, having the similar duration above and under the pins blacks is. So this is 12345 right here and under it. 12345 So this is the pawn gave lands drawn like this. Now we need to mark the foremost focal point in either side. Off this concave lands having a distance of 30 centimeter for which each hole is taken as one centimeters. So counting which 123456789 10 123456789 10. And finally 123456789 and 10. This is F one the primary concept center of attention off this concave lens. Now, the place we put the object at a distance of 60 centimeter approach simply double off this. So we increase the we prolong optic this principal axis a little more something. This is the predominant axes prolonged. Now we will count the position off this object. This is 123456789 10. Here, put this. This is 40 and 123456789 10 This is 50 and in any case 123456789 and 10. So the object is right here at this point whose height is just one centimeter reasonably small object. This is a B. Now we have to draw the image with the assist of Ray tracing. So we put the dimensions like this The first charge ranging from the top off the object and traveling parallel to the most important axes will pass up to the midpoint of this on cave lands and then can be diverged away such that on generating again it's going to appear toe come from this concept Focus first predominant focus. A standard task we have to do so is the line must pass thru F one and this point as much as the midpoint of this concave lens and arms. This is the blind. So this is the ray which will debate clear of them in Spalax is however on generating back. It should exactly come via this primary idea focus and this is the proper. So this is deviated away and on producing again. It is appearing toe come from this principle focus. So this is incidently and this is mirrored right now another race ranging from the similar level B. But this time it really pass through the optical heart. It should cross throughout the optical center and it will pass beneath deviated, so the dimensions will have to be kept ranging from level B. Then it should be debated until it starts passing through this point C like this. This is the precise position. So you'll draw in a different way. This is otherwise. Yep. Passing throughout the optical middle, which is passing Unterberg. Muted. Very shut. Raise a standard process. So these rays seemed to meet someplace at this level. Yeah, this is Mm. Here, like this. We can display it in any other colour. So this is the image. A dish beat us. And if we discover it is the stance, we count from the optical heart. 123456789 and 10. And this is 123456789 and 10. So, the use of that symbol, this stance comes out Toby, minus 20 centimeters. And if we look for the height off this symbol, this is approximately half the height off this block. So we can say it will be less than one centimeter or it is rather lower than the part off the block additionally. So this is something not up to 0.5 centimeter? No. We will to find each of those answers with the lend a hand off lens equation. So in the second phase of the problem the usage of lands equation, which says one through Q minus one by way of P is equivalent to 1 by f plugging in all recognized values. For me, this is 60 centimeters minus 60 for Memphis is minus 30. So one motorbike you seek the advice of Toby minus one via 30 minus one via 60 Taking L C M. This is 60. This is minus two minus one. So this is minus 3 by 60 or minus one via 20. So in the end the answer for this symbol is stressful comes out to B minus 20 centimeter, which is precisely same which now we have discovered with the help of repressing Now toe to find the height off symbol the use of the expression for linear magnification. Chemicals toe Q through Peter A. Show off distance off finish. So the distance off object or H two through H. One top off end to the height off object plugging in unknown values for you. This is minus 20 for P. This waas minus 60 is equal to H two, which is missing for H one this waas one Santa meet up for canceling all these issues. The value of this H two comes out Toby plus one through three centimeters or 1/3 centimeter. Which is also precisely identical as we now have discovered in the first phase of the problem with the lend a hand off Really crazy. Thank you.
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