Reflection and Mirrors Legacy Problem #13 Guided Solution

Here in this lab, a couple of students have a concave mirror of known focal length, F equal 32.0 centimeters. They are varying the object distances and measuring the image distances. And ironically enough, knowing the focal length and the object distance, the actual image distances could be predicted using the mirror equation. The mirror equation states that the reciprocal of the object distance plus the reciprocal of the image distance equal the reciprocal of the focal length. In symbol form, that's 1 over dou plus 1 over di equal 1 over F. In each of these three parts of this problem, we know the dou value and the F value. The F is 32.0 centimeters and in part A, the dou value is 85.3 centimeters. I can rearrange my mirror equation to the form of 1 over di equal 1 over F minus 1 over dou. And for part A, I substitute in 32.0 for F and 85.3 centimeters for dou. And when I evaluate the right side of the equation, I end up getting 0.019527. That's not the image distance. That's the reciprocal of the image distance. So if I take the reciprocal of the image distance, I get di and I'd have to do the same thing for the right side of the equation, the reciprocal of 0.019527 comes out to be 51.2120 centimeters. That's the image distance for part A. I can round that to three significant digits, 51.2 centimeters. For part B, I simply substitute in dou value of 64 such that 1 over 32 minus 1 over 64 becomes 0.015625 and I can reciprocate that and I get 64.0 centimeters and that's again the image distance for part B. For part C, I do much the same thing, 1 divided by 32 minus 1 divided by 48.1, which evaluates to 0.010460 and take the reciprocal of that number comes out to be the image distance of 95.6025 centimeters. I can round that to three significant digits, 95.6 centimeters.