Answer to Question #348865 in General Chemistry for Ashter

Question

a flea is located 3.0 cm from a convex spherical surface of radius 10 cm. Where is the image of the flea?

Answer 1

Hello everyone. Let's proceed to solve this question. So in the first part we're given a convex mirror. So this is a convex mirror here and it is polished this side and we have given the radius of curvature for this mirror are to be 10 cm. So the focal length will be five cm. So we have our principal axis here. So this is the principal axis. And from sign convention we know that this side is taken as negative and this side is positive. So we have a flea at a distance of so focus is somewhere here. So the focus is somewhere here. F. And we have a flea which is at a distance of 3cm. So this is The flea here which is at a distance of three cm from the mirror. Alright, so we have our object distance you to we have to take the sign convention also. So negative three cm. Alright, so we need to find where the image of this fleet will be formed. So we'll use the middle formula. So we'll use the middle formula. So from the middle formula. Sorry. So from the middle formula we know that one by U. Plus one by V is equal to one by F. Where we is the image distance from the mirror. So let's substitute the values. So one by negative of three Plus one by V is simply equal to one by five. So on simplifying this movie comes out to be 15 x eight cm and this is equal to 1.875 cm. So the image will form somewhere here as we comes out to be positive. So this is the image of the field. All right. So let's see the second part of the question. So in the second part of the question we have been given a concave mirror. So let's draw a concave mirror. So a concave mirror is somewhat looks like this and it is polished this side. All right. Again we have a principal access here. Okay Now here the image distance has given us to be so this side is again negative. This side is positive. So it's given us to be two cm. Focus the radius of curvature was eight cm. focus lies here. So we'll take the focus to be negative of four cm. As focused is simply the radius of curvature divided by two. Alright, so let's again use the middle formula. So one by V. Plus one by U. Is equal to one by F. So let's substitute the values. So one by V plus negative of one by U. is equal to Negative of one x 4. All it So we comes out to be simplifying one x 4 which is 0.025. Sorry 0.25 0125 cm. So as we comes positive. So the image will form somewhere here At a distance of 0.25 from the mirror. Now we need to find the size of that image. Alright, so magnification. We know negative of V by you and that's equal to the size of image by the size of the object. Now let's substitute the values here. So negative of 0.25 divided by negative of two is equal to the size of image by the size of object. So this comes out to be 0.125. That's equal to the size of image by size of object. And finally the size of image is 0.125. Multiplied by the size of object. So as we can see the image is 0.125. Multiplied by the size of objects. So the image formed will be magnified. So this is the final answer. I hope you understood it. Thank you.