IIT-JEE 2010 - Career Point

advice should be followed when dealing with
lenses in the following sections.)
It is usually best to orient your diagrams
consistently with the incoming rays traveling
from left to right. Don't draw a lot of other rays at
random ; stick with the principal rays, the ones
you know something about. Use a ruler and
measure distance carefully ! A freehand sketch
will not give good results.
If your principal rays don't converge at a real
image point, you may have to extend them
straight backward to locate a virtual image point,
as figure (b). We recommend drawing the
extensions with broken lines. Another useful aid
is to color-code the different principal rays, as is
done in figure(a) & (b).
Q
P
Q
P
4
3
I
2
4
1
4
2 2
2
1
(b)
4
C
1
3
(a)
v
P'
Q'
F
Q'
3
P' F
1 1 1
Check your results using Eq. + = and the
s s' f
y'
s'
magnification equation m = = − . The
y s
results you find using this equation must be
consistent with your principal-ray diagram; if not,
double-check both your calculation and your
diagram.
Pay careful attention to signs on object and image
distances, radii or curvature, and object and image
heights. A negative sign on any of these
quantities always has significance. Use the
equations and the sign rules carefully and
consistently, and they will tell you the truth !
Note that the same sign rules (given in section)
work for all four cases in this chapter : reflection
and refraction from plane and spherical surfaces.
Step 4: Evaluate your answer : You've already
checked your results by using both diagrams and
equations. But it always helps to take a look back
and ask yourself. "Do these results make sense ?".
v
C
1. How will you arrange the two mirrors so that
whatever may be the angle of incidence, the incident
ray and the reflected ray from the two mirrors will be
parallel to each other.
Sol.
B
A
Solved Examples
θ
P
Q
i 1 i 2
i 1 i 2
The total deviation of the ray is given by
δ =180 – 2i 1 + 180 – 2i 2
= 360 – 2(i 1 + i 2 )
For the resultant ray to be parallel, δ should be 180º
∴ 360 – 2(i 1 + i 2 ) = 180
i.e., i 1 + i 2 = 90º
From the geometry of the figure i 1 + i 2 = θ
θ Angle between the mirrors should be 90º.
2. Rays of light strike a horizontal plane mirror at an
angle of 45º. At what angle should a second plane
mirror be placed in order that the reflected ray finally
be reflected horizontally from the second mirror.
Sol. The situation is shown in figure
C G
A
S θ θ
D
P
45º
45º
B
N
Q
The ray AB strikes the first plane mirror PQ at an
angle of 45º. Now, we suppose that the second
mirror SG is arranged such that the ray BC after
reflection from this mirror is horizontal.
From the figure we see that emergent ray CD is
parallel to PQ and BC is a line intersecting these
parallel lines.
So, ∠DCE = ∠CBQ = 180º
∠DCN + ∠NCB + ∠CBQ = 180º
θ + θ + 45º = 180º ∴ θ = 67.5º
As ∠NCS = 90º, therefore the second mirror should
be inclined to the horizontal at an angle 22.5º.
3. An object is placed exactly midway between a
concave mirror of radius of curvature 40 cm and a
convex mirror of radius of curvature 30 cm. The
mirrors face each other and are 50 cm apart.
Determine the nature and position of the image
formed by the successive reflections, first at the
concave mirror and then at the convex mirror.
C
XtraEdge for IIT-JEE 28 DECEMBER 2009