In: Physics
ESSAY PART:
A group of students of students wants to build an apparatus that produces a mirage of an object. For this purpose, they need to know the radius of curvature of the concave mirror they are using. Please explain how they can achieve their goal? Make sure to indicate variables, equipment necessary and possible procedure. (10pts) Also on the space labeled Experimental setup sketch your apparatus.
Experimental Setup: (5 pts)
Distance to the object (cm) |
10 |
15 |
20 |
25 |
30 |
35 |
Distance to the image (m) |
30 |
15 |
12 |
10.7 |
10 |
9.5 |
Complete the above table by manipulating the data so that you are able to determine the focal length of the mirror. (20pts)
Mathematical Equation: ____________________ Meaning of the Slope: Meaning of the y-intercept:
Distance to Object u (cm) | 10 | 15 | 20 | 25 | 30 | 35 |
Distance to image v ( cm ) | 30 | 15 | 12 | 10.7 | 10 | 9.5 |
Reciprocal of distance to Object , 1/u ( cm-1) | 0.1 | 0.067 | 0.05 | 0.04 | 0.033 | 0.029 |
Reciprocal of distance to Image , 1/v ( cm-1) | 0.033 | 0.067 | 0.083 | 0.093 | 0.1 | 0.105 |
We have mirror equation :- (1/v) + (1/u) = (1/f)
where f is focal length of mirror
Hence get, (1/v) = -(1/u) + (1/f) ................(1)
From eqn.(1), we see that if we use (1/v) as ordinate ( y-variable) and (1/u) as abscissa ( x-variable)
we have linear equation.
Slope of graph of linear equation is -1 and y intercept of graph of linear equation is
reciprocal of focal length of concave mirror.
Radius of curvature of concave mirror = twice of focal length
Graph shown above is plotted using excel using (1/u) and (1/v) values
Trendline equation ( linear ) is shown in the graph.
y-intercept = reciprocal of focal length 1/f = 0.133 cm-1
focal length = 1/0.133 = 7.52 cm
Radius of curvature = 15.04 cm
If actual radius of curvature is 14.8 cm , then % error = [ (15.04 -14.8) / 14.8 ] 100 1.61