Question

As shown in the figure, a particle is placed at O in front of a plane mirror M. A man at P can move along path $PY$ and $PY'$. Then which of the following is true?

• A. For all points on $PY$, man can see the image of O
• B. For all points on $PY'$, man can see the image, but for no point on $PY$, he can see image of O
• C. For all points on $PY'$, he can see the image but on $PY$, he can see the image only upto distance d
• D. He can see the image only upto a distance d on either side of P

Answer 1

Answer: (A)

A

Answer 2

To determine where the man can see the image of particle O, we need to find the field of view of the mirror from the observer's perspective.

1. Establish the Coordinate System and Positions:

   Let the point P be the origin (0,0).
   The path PY is along the positive y-axis, so Y is at (0,d).
   The path PY' is along the negative y-axis.
   The figure shows the particle O at (0,0).
   The plane mirror M is located at a distance 'd' from O, along the negative x-axis. Its height is also 'd'.
   So, the mirror M is a vertical line segment from (-d, 0) to (-d, d). Let's call the bottom edge $M_1 = (-d, 0)$ and the top edge $M_2 = (-d, d)$.

2. Locate the Image of O (O'):

   For a plane mirror, the image is formed as far behind the mirror as the object is in front of it.
   Object O is at (0,0). The mirror is along the line x = -d.
   The distance of O from the mirror is $|0 - (-d)| = d$.
   So, the image O' will be at x = -d - d = -2d. The y-coordinate remains the same.
   Therefore, the image O' is at (-2d, 0).

3. Determine the Field of View:

   The field of view is the region from which an observer can see the image. This region is bounded by the rays originating from the image and passing through the extreme edges of the mirror.

   • Ray 1: From the image O'(-2d, 0) to the bottom edge of the mirror $M_1(-d, 0)$.
     This ray lies along the line y = 0.

   • Ray 2: From the image O'(-2d, 0) to the top edge of the mirror $M_2(-d, d)$.
     The slope of this line is $m = \frac{d - 0}{-d - (-2d)} = \frac{d}{d} = 1$.
     Using the point-slope form ($y - y_1 = m(x - x_1)$) with O'(-2d, 0):
     $y - 0 = 1(x - (-2d))$
     $y = x + 2d$.

   The observer (man) is moving along the y-axis, which means his x-coordinate is 0. To find the range of y-coordinates where the man can see the image, we substitute x=0 into the equations of the two rays:

   • For Ray 1 (y = 0): At x=0, y = 0.
   • For Ray 2 (y = x + 2d): At x=0, y = 0 + 2d = 2d.

   So, the man can see the image when his y-coordinate is between 0 and 2d, inclusive. That is, $0 \le y \le 2d$.

4. Analyze the Man's Paths:

   • Path PY: This path starts at P(0,0) and goes upwards to Y(0,d).
     Along this path, the y-coordinates range from 0 to d ($0 \le y \le d$).
     Since this entire range ($0 \le y \le d$) falls within the visibility range ($0 \le y \le 2d$), the man can see the image for all points on path PY.

   • Path PY': This path starts at P(0,0) and goes downwards (negative y-axis).
     Along this path, the y-coordinates are negative ($y < 0$).
     Since this range ($y < 0$) does not overlap with the visibility range ($0 \le y \le 2d$), the man cannot see the image for any point on path PY'.

5. Evaluate the Options:

   • (A) For all points on PY, man can see the image of O. (TRUE, as $0 \le y \le d$ is within $0 \le y \le 2d$).
   • (B) For all points on PY', man can see the image, but for no point on PY, he can see image of O. (FALSE, as he cannot see on PY').
   • (C) For all points on PY', he can see the image but on PY, he can see the image only upto distance d. (FALSE, as he cannot see on PY').
   • (D) He can see the image only upto a distance d on either side of P. (FALSE, as he can see up to 2d on the positive y-axis and not at all on the negative y-axis).