The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (b) If the object is 1.00 mm high, what is the height of its formed by the magnifier?

Question

The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (b) If the object is 1.00 mm high, what is the height of its  formed by the magnifier?

Accepted Answer

Hi everyone in this practice problem, we're being asked to determine the height of an image formed by a loop of a watchmaker. We'll have a watchmaker's loop with a focal length of 4.5 centimeter. Used to magnify the minute wheel of a watch placed 3. centimeters in front of the magnifier. The minute wheel is two millimeter tall and we're being asked to determine the height of the image formed by the loop. The options given are a 2.34 millimeter B 3.71 millimeter C 6.91 millimeter and lastly D 13.8 millimeter. So in this particular practice problem, the size image or the height of the image, which is going to be the absolute value of Y prime can be calculated by multiplying M or the magnification with Y where Y is the actual height. We want to first find what the M is. So we want to employ or use the object image relationship for a lens to calculate the position of the image or S prime. The object image relationship is one divided by S plus one, divided by S prime equals to one divided by F. What we're interested to find is the S prime or the image distance form with the loop. The image position as prime using this formula can be obtained through some rearrangement where S prime will then equals to S multiplied by F divided by S minus F. So we know what the S or the uh actual object distance is and we know what the F is, which is the focal length. So you want to substitute all the dose information into this formula. So first, the S or the image distance is going to be um the 3.85 centimeter here. So we have 3.85 centimeter multiplied by F which is going to be 4.5 centimeter divided by S which is 3.85 centimeter minus F which is 4.5 centimeter. So calculating all of this, this will give us the image distance or S prime to then be equals to negative 26.65 centimeter. Now we want to calculate the magnification or M where M will be equals to negative prime divided by S. So let's substitute that information in. So M will be equals to negative of S prime which is negative 26.65 centimeter divided by S which is going to be 3.85 centimeter. Calculating this, we have the magnification to then be equals to 6.922. Awesome. So now, we can actually pull everything together into the actual uh formula that we're gonna use for the height of the image formed by the loop where Y prime will be equals to the absolute value of M multiplied by Y. The absolute value of M will be 6.922 multiplied that by Y which is a two millimeter calculating this, we will get the actual or the image height of the I the image height form by the loop which is going to equals to 13.8 millimeter. So 13.8 millimeter will be the answer to this particular practice problem that which will correspond to option D in our answer choices. So the height of the minute wheels image is going to be 13. millimeter formed by the loop. So that'll be it for this video. If you guys still have any sort of confusion, just please make sure to check out our and videos on our website. But other than that, that will be it for this video. Thank you.

The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (b) If the object is 1.00 mm high, what is the height of its formed by the magnifier?

Verified Solution

Key Concepts

Focal Length

Magnification

Thin Lens Formula