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
10.2k views
in Physics by (46.1k points)
closed by

A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 18 cm at a distance of 12 cm from it. Use lens formula to determine the position, size and nature to the image formed.

1 Answer

+1 vote
by (49.2k points)
selected by
 
Best answer

According to the question; 

Object distance = -12cm; 

Image distance = v; 

Focal length = 18cm; 

By lens formula;

\(\frac1v-\frac1u=\frac1f\)

⇒ \(\frac1v-\frac1{-12}=\frac1{18}\)

⇒ \(\frac1v=\frac1{18}-\frac1{12}\)

⇒ \(\frac1v=\frac{2-3}{36}-\frac{-1}{36}\)

⇒ \(\frac1v=\frac{-1}{36}\)

⇒ v = -36cm. 

Therefore, image is formed at 36 cm in front of lens. 

Now; 

Height of object h1= 5cm;

Magnification = \(\frac{h_2}{h_1}=\frac{v}u\)

Putting values of v and u 

Magnification = \(\frac{h_2}{5}=\frac{-36}{-12}\)

⇒ \(\frac{h_2}{5}=3;\)

⇒ h2 = 3 x 5 = 15

Height of image is 15 cm. 

Thus, the image is virtual, enlarged, and erect and three times the size of object.

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

...