A lightbulb is 3.0 m from a wall. What are the focal length and t...

Question

A lightbulb is 3.0 m from a wall. What are the focal length and the position (measured from the bulb) of a lens that will form an  on the wall that is twice the size of the lightbulb?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A science instructor holds her phone 2.0 m away from a whiteboard. The teacher wants to magnify the device's screen three times using a lens, determine the focal length and distance at which the lens should be positioned with respect to the phone to obtain this effect. So that's our end goal is we're trying to figure out what the focal length and distance must be in order that the lens should be positioned with respect to the phone in order to obtain this magnification of three times, that's our final answers that we're trying to solve for our end goal or what we're ultimately trying to solve for is putting it simply, we just need to figure out the focal length and the distance at which the lens needs to be positioned in order to obtain the screen magnification of three times. Awesome, we're also given some multiple choice answers. Let's read them off to see what our final answer pair might be. So A is the lens should be 50 centimeters from the phone and have a focal length of 38 centimeters. B is, the lens should be 50 centimeters from the phone and have a focal length of 72 centimeters. C is the lens should be 100 centimeters from the phone and have a focal length of 38 centimeters and D is, the lens should be 100 centimeters from the phone and have a focal length of 72 centimeters. Oh That was a mouthful. So first off, let us solve for the sum of the image distance which is gonna be denoted as S prime. And the object distance denoted as S which is equal to 2.0 m which we can write as and let's call this equation one. So S prime plus S is equal to 2.0 m. So now at this stage, we must recall and use the equation for the magnification which we're gonna denote as M. So note that the magnification can be either positive or negative three and can be calculated as follows. So M equals so the magnification M equals negative S prime divided by S and let's call this equation two. So by using equations one and two, we can determine the image distance S prime as follows. So negative S prime divided by S is equal to plus or minus three where once again to reiterate three is the magnification. So rearranging the isolate and sulfur S prime, we will find that S prime is equal to plus or minus three, multiplied by S which we can take it even further and write plus or minus three S3 multiplied by S plus S is equal to 2.0 m. So continuing to separate like so now what we're trying to do is we're trying to take this equation, we're trying to isolate it and solve for S by itself. So to do that, we will get four S equals 2.0 m which simplifying once again, you'll get that S is equal to 0.5 m or because you have to remember it's plus or minus or negative two multiplied by S is equal to 2.0 m which s in that case will equal negative 1.0 m. So you basically, you're using algebra to isolate and rearrange just a sulfur S by itself. But we have to remember we have the plus and the minus case. Awesome. So with that in mind, we need to note that and I'll fox it in red. We need to note that S equals negative 1.0 m will lead to an un physicals object distance. Thus, this solution is not physically acceptable. So it's not correct. So therefore, s and we'll, so therefore, we need to note that S must equal 0.5 m or 50 centimeters. So that is our distance that it has to be at and we'll box it in green. So that is the correct answer for the object distance. Awesome. So, so now that we know that S must equal 0.5 m or 50 centimeters and the magnification will be negative. So this is important. So the magnification in this case will be negative, thus, we can write that negative three is equal to negative S prime divided by S which means that S prime is equal to three S which is equal to 1.5 m. Now, at this point, we need to recall and use the thin lens equation to help us calculate the focal length. And that states that one divided by S plus one divided by S prime is equal to one divided by F. So we need to rearrange this equation to isolate and sulfur F. So using a little bit of algebra, we will find that F is equal to S multiplied by S prime divided by S plus S prime, which now we can plug in our known variables and solve for the focal length. So we know that S is 0.5 m and we need to multiply it by our S prime value which is equal to 1.5 m all divided by 0.5 m plus 1.5 m. So when we plug that into our calculator, we will get 0.375 m. But as we can see all of our multiple choice answers are in centimeter value. So we need to multiply our final answer by 100. And when we do that, we will get 37.5 centimeters which will be approximately equal to 38 centimeters. And that is our final answer for the focal length. Awesome, we did it. So looking at our multiple choice answers, the correct answer has to be the letter A which states that the lens should be 50 centimeters from the phone and have a focal length of 38 centimeters. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

A lightbulb is 3.0 m from a wall. What are the focal length and the position (measured from the bulb) of a lens that will form an on the wall that is twice the size of the lightbulb?

Watch next

Thin Lens And Lens Maker Equations practice set