Solving Mirror Equation Questions: Step-by-Step Guide

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SUMMARY

The discussion focuses on solving mirror equation problems related to concave mirrors. The first question involves finding the image location, height, and magnification for an object placed at the center of curvature, utilizing the relationship that the object distance (d) equals twice the focal length (f). The second question applies the mirror equation (1/f = 1/do + 1/di) and magnification equation (hi/ho = -di/do) to determine the image properties for an object 1.5 m from a concave mirror with a 50 cm focal length, resulting in an image height of approximately 2.576 cm.

PREREQUISITES
  • Understanding of the mirror equation (1/f = 1/do + 1/di)
  • Knowledge of magnification equation (hi/ho = -di/do)
  • Familiarity with concave mirror properties and focal length concepts
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the derivation and applications of the mirror equation in optics
  • Explore the concept of focal length and its significance in concave mirrors
  • Practice solving various problems involving magnification and image properties
  • Learn about real vs. virtual images in concave mirrors and their characteristics
USEFUL FOR

Students studying optics, physics educators, and anyone seeking to understand the principles of concave mirrors and their applications in real-world scenarios.

rhxoehwhfh
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I still find these stuff hard, I have 2 question to ask about...

1. Use the mirror equation to find the image location and the height of an object placed at the centre of curvature of a concave mirror. Also find the magnification. Hint: What is the relation between the focal length and the object distance, d, for this situation?

2. Using the mirror equation and the magnification equation, find the four properties of the image formed in a concave mirror with a focal length of 50 cm, if the object is 1.5 m from the mirror and is 2.5 cm high.


Can anyone help me and show step by step how to solve these questions?,,, I really am having a difficulties... please..
 
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for second question, i tried and got this
. Mirror equation
1/f = 1/do + 1/di
1/50= 1/1.5 + 1/di
di= -1.546 cm do= 1.5cm
Magnification equation
hi/ho = - di/do
hi/2.5 = -1.546/ 1.5
hi =2.576 cm ho= 2.5cm
is this right?
 
for #1, the center of curvature is 2 x the focal length so you can say d = 2f.
Now plug that into the 1/f = 1/d + 1/di

For magnification, use the equation you provided.
 

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