This section includes many important IIT JEE Physics problems in Ray Optics topics that are considered to be very important from IITJEE exam point of view. Here the solutions for these problems are very detailed with explanation for each step. This section is very helpful for students to get very high scores in IIT JEE.
IIT JEE Physics Problems in Concave Mirror topic in Ray Optics concepts And Their Solutions with Detailed Step by Step Explanation.
Ray Optics – Concave Mirror
🟩 Problem 1:
A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is -3, then the magnitude of the radius of curvature of the mirror is:
Options:
(1) 3.75 cm (2) 30 cm (3) 7.5 cm (4) 15 cm
✅ Solution:
📊 View Ray Diagram
Step 1: Use magnification formula
Magnification for mirrors is given by:
\( m = \frac{-v}{u} \)
- \( m \) = magnification
- \( v \) = image distance (from mirror pole)
- \( u \) = object distance (negative for real object)
Given: \( m = -3 \Rightarrow -3 = \frac{-v}{u} \Rightarrow v = 3u \tag{1} \)
Step 2: Use object-image distance
Given: \( |v – u| = 20 \text{ cm} \tag{2} \)
From (1): \( v = 3u \), substitute into (2):
\( |3u – u| = 20 \Rightarrow 2u = 20 \Rightarrow u = \pm10 \)
Since object is real: \( u = -10 \text{ cm} \), so \( v = 3u = -30 \text{ cm} \)
Step 3: Use mirror formula
Mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
Substitute:
\( \frac{1}{f} = \frac{1}{-30} + \frac{1}{-10} = -\left( \frac{1}{30} + \frac{1}{10} \right) = -\frac{4}{30} \Rightarrow f = -7.5 \text{ cm} \)
Step 4: Find Radius of Curvature
\( R = 2f = 2 \times (-7.5) = -15 \text{ cm} \)
Magnitude: \( |R| = \boxed{15 \text{ cm}} \)
✅ Final Answer:
Radius of Curvature is :
(4) 15 cm🟩 Problem 2:
A point light source lies on the principle axis of concave spherical mirror with radius of curvature 160cm. Its image appears to be back of the mirror at a distance of 70cm from mirror. What is the location of the light source?
Options:
(1) 3.73 cm(2) 27.3 cm (3) -37.3 cm (4) -2.73 cm
✅ Solution:
Step 1: Use mirror formula
Mirror formula is given by:
\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
(or)
\(\frac{2}{R} = \frac{1}{v} + \frac{1}{u} \)
- \( f \) = focal length of the mirror
- \( R \) = radius of curvature of the mirror
- \( v \) = image distance (from mirror pole)
- \( u \) = object distance (negative for real object)
Step 2: Apply mirror formula
Mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
Substitute:
\( \frac{2}{-160} = \frac{1}{u} + \frac{1}{70} \Rightarrow \frac{1}{u} = -\frac{2}{160} -\frac{1}{70} = -\left( \frac{2}{160} + \frac{1}{70} \right) = -\frac{15}{560}\)
Step 3: Find distance of Light Source
\( u = -\frac{560}{15} \Rightarrow u = -37.3 \text{ cm} \)
Magnitude of Object Distance: \( |R| = \boxed{37.3 \text{ cm}} \)