What should be the distance between the object in Exercise 9.30 and the magnifying glass

+1 vote
25 views
asked Jan 8, 2018 in Physics by sforrest072 (157,439 points) 60 409 932
edited Mar 5, 2018 by Vikash Kumar

What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2? Would you be able to see the squares distinctly with your eyes very close to the magnifier?

 [Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

1 Answer

+1 vote
answered Jan 8, 2018 by mdsamim (213,225 points) 5 10 15
edited Mar 5, 2018 by Vikash Kumar
 
Best answer

Area of the virtual image of each square, A = 6.25 mm

Area of each square, A0 = 1 mm2 

Hence, the linear magnification of the object can be calculated as:

Focal length of the magnifying glass, f = 10 cm 

According to the lens formula, we have the relation:

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

One Thought Forever

“There is a close connection between getting up in the world and getting up in the morning.“
– Anon
~~~*****~~~

...