Calculate the field of view of a point object placed symmetrically at a distance 's' from a plane mirror of length 'l', (consider only the region between the mirror and a line parallel to the mirror at distance 's').
Calculate the field of view of a point object placed symmetrically at a distance 's' from a plane mirror of length 'l', (consider only the region between the mirror and a line parallel to the mirror at distance 's').
The correct option is C 3ls2
Remember, field of view is the area bounded by the rays reflected from the edges/extremities of the mirror. Further, in this question you have been asked to consider only the area bounded by these rays and the line parallel to the mirror, at a distance s.
To find the field of view keep only these 2 things in mind:
1.What point object is the source of the rays?
2.What do the reflected rays (keep in mind the laws of reflection) from the edges/ extremities of the mirror look like.
After this all that remains is the simple application of geometry (mostly congruency or similarity of triangles). Let us get started then!
In this case the point object is placed symmetrically 's' from a plane mirror of length 'l', so the ray diagram representing field of view is given as:
(read between the lines tip: by 'symmetrically' we mean that the objects is placed such that it is equidistant from either ends of the mirror)
Now, the area between the two rays and a line passing through O will be the required field of view is the area of the trapezium CBDE below: (BF and DG are normal)
Now, to find the area of trapezium (recall the formula,area=sum of parallel sides x perpendicular distance between them2)we need the length of the parallel sides CE and BD, and the perpendicular distance AO between them. Directly from the figure we see that BD= l and AO= s, so 2/3rd of our work is done!
To find CE, we can do the following: OFBA is a rectangle, so, OF=AB=l2. Similarly, OG=AD=l2. Now, the right triangles OFB and CFB are congruent, as the side FB is common and the angles ∠OBF=∠CBF. Therefore CF=OF=l2. Similarly, GE=OG=l2. Therefore, our last part of the puzzle, CE=CF+OF+OG+GE=4×l2=2l.
Finally the area = (CE+BD)xAO2=(2l+l)x82=3ls2.
Remember, field of view is the area bounded by the rays reflected from the edges/extremities of the mirror. Further, in this question you have been asked to consider only the area bounded by these rays and the line parallel to the mirror, at a distance s.
To find the field of view keep only these 2 things in mind:
1.What point object is the source of the rays?
2.What do the reflected rays (keep in mind the laws of reflection) from the edges/ extremities of the mirror look like.
After this all that remains is the simple application of geometry (mostly congruency or similarity of triangles). Let us get started then!
In this case the point object is placed symmetrically 's' from a plane mirror of length 'l', so the ray diagram representing field of view is given as:
(read between the lines tip: by 'symmetrically' we mean that the objects is placed such that it is equidistant from either ends of the mirror)
Now, the area between the two rays and a line passing through O will be the required field of view is the area of the trapezium CBDE below: (BF and DG are normal)
Now, to find the area of trapezium (recall the formula,area=sum of parallel sides x perpendicular distance between them2)we need the length of the parallel sides CE and BD, and the perpendicular distance AO between them. Directly from the figure we see that BD= l and AO= s, so 2/3rd of our work is done!
To find CE, we can do the following: OFBA is a rectangle, so, OF=AB=l2. Similarly, OG=AD=l2. Now, the right triangles OFB and CFB are congruent, as the side FB is common and the angles ∠OBF=∠CBF. Therefore CF=OF=l2. Similarly, GE=OG=l2. Therefore, our last part of the puzzle, CE=CF+OF+OG+GE=4×l2=2l.
Finally the area = (CE+BD)xAO2=(2l+l)x82=3ls2.
