For each thin lens shown in Fig. E34.37, calculate the location o...

Question

For each thin lens shown in Fig. E34.37, calculate the location of the  of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes. (b)

Accepted Answer

Hi everyone in this practice problem, we will have two types of thin lenses, converging and diverging. And we're being asked for each lens to find how far from the lens the image of an object is which is placed 22 centimeters to the left of the lens. The index of refraction of the lens of material is 1.43. And the options listing the different S prime or the distance for the image of an object from the lens. So it is given in the problem statement or given in the figure that we have double cons and double concave lenses and the r of each lenses are listed in the figure as well. And we want to note or recall that the sine convention for the optical radius of curvature is as follows. So if the vertex lies to the left of the center of curvature, then the radius is positive. And if the vertex lies to the right of the center of curvature, then the radius of the curvature is going to be negative. First, we're going to look at the double convex uh lens and with the double convex lens, we're going to use the lens maker equation to calculate the focal length for the lens. So we know that the lens maker equation, if we recall is going to be represented by one divided by F equals to in parentheses and minus one close parenthesis, open parenthesis, one divided by R one minus one, divided by R two, close parentheses. We wanna plug in all the values for the double con facts to get one divided by F or F. So you will get one divided by F equals to N is the index of reflection for the less material which is 1.43 minus one multiplied by open parentheses. One divided by R one is going to be given by centimeter minus R two is going to be given by negative 15 centimeter based on the uh radius sine con convention that we just determined previously. And in calculating this, this will equal to 0.43 multiplied by uh three divided by 20 centimeter. And then all of this calculating all of this, we will then get the F to then be equal to 15.5 centimeter just like. So now uh that we have obtained the focal line, we can calculate the image distance and the image distance. We want to recall the object image relationship where one divided by S uh plus one divided by S prime will equals to one divided by F. And in this case, we are interested to find S prime. So rearranging the equation to get S prime, we will get S prime to be equal to s multiplied by F divided by S minus F. And then plugging in all the values we can then get uh as to be equal to 15.5 centimeter which is the focal length that we just obtained. And then the object distance, which is uh given in the problem statement to be 22 centimeter divide that with 22 centimeter minus 15.5 centimeter that will give the S prime value of 52.5 C meter. So the image is going to be 52.5 centimeter to the right of the lens just like. So that's for the double convex one. So now let's look into the double concave lens where we are going to do pretty much the same uh steps here. So first, the double con cave, we're going to use the um lens maker equation where one divided by F equals to N minus one multiplied by one divided by R one minus one, divided by R two. And then we're going to find the focal length where one divided by F will be equal to one minus 41.43 minus one multiplied by um one divided by R one is going to be negative 12 centimeter minus one divided by 16 centimeter based on the radius confine that we just gave this will be equal to 0.43 multiplied by negative seven divided by centimeter. And then that will give the focal length of negative 15.95 centime meter. Now, for the image distance we're going to use exactly the same thing and exactly the same formula. So S prime will be equal to SFM uh divided by S minus F. And then we will uh substitute all of the values that we have where we have to be equal to centimeter F is negative 15.95 centimeter which we just obtained divided by 22 centimeter minus negative 15.95 centimeter. And then calculating this, this will give the S prime value of negative 9. centimeter or essentially S prime is going to be equal to 9.2 centimeter to the left of lens just like so awesome. So there we have the answer for the double concave the image uh of the object is going to be 9.2 centimeter to the left of the lens. And then for the double con uh fax lens, the image of the object is going to be 52.5 centimeter to the right of the lens. And all of this will correspond to option C in our answer choices. So option C will be the answer to this particular practice problem. And that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

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For each thin lens shown in Fig. E34.37, calculate the location of the of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes. (b)

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