question_answer

For a concave mirror, if real image is formed the graph between \[\frac{1}{u}\] and \[\frac{1}{v}\] is of the form

A)                                    

B)                            

C)                                    

D)

Correct Answer: A

Solution :

                   Since \[\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\] Þ \[\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}\] Putting the sign convention properly \[\frac{1}{(-v)}=\frac{-1}{(-u)}+\frac{1}{(-f)}\] Þ \[\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}\] Comparing this equation with y = mx + c Slope = m = tanq = ? 1 Þ q = 135° or ? 45° and intercept \[C=+\frac{1}{f}\]