Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
6.5k views
in Physics by (80.8k points)

A thin converging lens L of focal length 10.0 cm is used as a simple magnifier to examine an object O that is placed 6.0 cm from the lens.

(a) On the figure above, draw a ray diagram showing at least two incident rays and the position and size of the image formed.

(b) i. Indicate whether the image is real or virtual.

____ Real ____ Virtual

ii. Justify your answer.

(c) Calculate the distance of the image from the center of the lens. (Do NOT simply measure your ray diagram.)

(d) The object is now moved 3.0 cm to the right, as shown above. How does the height of the new image compare with that of the previous image?

____ It is larger. ____ It is smaller. ____ It is the same size. Justify your answer.

1 Answer

+1 vote
by (52.5k points)
selected by
 
Best answer

(a)

(b) i&ii. This image is virtual because it is on the same side as the object and is not projectable, this can also be proven mathematically as shown below.

(c) 1/f = 1/do + 1/di … 1/10 = 1/6 + 1/di … di = –15 cm

(d) This image is very close to the focal point. The rays will intersect further away and make a much larger image as shown in the ray diagram. This could also be proved mathematically comparing the magnification before (2.5) to the magnification after (10).

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.

Categories

...