A) 5 cm
B) varies from 5 to 7.5 cm
C) cannot be found
D) varies from 7.5 to 9 cm
Correct Answer: B
Solution :
[b]| Image of the farther edge of the phone will be at centre of curvature itself of the same size -5 cm. But the other edge at 50 cm will form its image at v satisfying, \[\frac{1}{-30}=\frac{1}{\upsilon }+\frac{1}{-50}\Rightarrow \,\frac{1}{\upsilon }=\frac{1}{50}-\frac{1}{30}=\frac{3-5}{150}\] and \[\upsilon =-75\,cm\] |
| Image height for object distance 50 cm is, |
| \[{{h}_{I}}={{h}_{0}}\times \left( -\frac{\upsilon }{u} \right)=w\times \left( -\frac{\upsilon }{u} \right)\] |
| \[=5\times \left( -\frac{(-75)}{(-50)} \right)=-7.5\,cm\] |
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