Q68.A light ray emerging from the point source placed at P(1, 3) is reflected at a point Q in the axis of x. If the reflected ray passes through the point R (6, 7), then the abscissa of Q is: (1) 1 (2) 3 (3) 7 (4) 5 2 2
What This Question Tests
This question applies the concept of reflection of a point across the x-axis and uses the property that the incident ray, the normal at the point of reflection, and the reflected ray are coplanar, or more simply, the reflected point and the original point lie on opposite sides of the reflecting line such that the line connecting the reflected point and the target point is straight.
Concepts Tested
Formulas Used
Reflected point of (x,y) in x-axis is (x,-y)
Slope of a line (y2-y1)/(x2-x1)
📚 NCERT Sections This Tests
10.2 — What Is The Shape Of The Wavefront In Each Of The Following Cases:
Physics Class 12 · Chapter 10
10.2 What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.
9.8 — A Beam Of Light Converges At A Point P. Now A Lens Is Placed In The
Physics Class 12 · Chapter 9
9.8 A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
9.17 — (A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection
Physics Class 12 · Chapter 9
9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.
📋 Question Details
- Chapter
- Coordinate Geometry
- Topic
- Reflection of a point
- Year
- 2013
- Shift
- 09 Apr Online
- Q Number
- Q68
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 10: Straight Lines
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