Q: Focal length of convex mirror M1 is 20 cm and that of concave mirror M2 is 30 cm. Find position of second image I2. Take first reflection from M1.

For M1 Using the mirror formula, u1=−20 cmf1=+20 cm Solving, we get For M2 v1+u1=f1, we have v11+−201v1u2f2=+201=+10 cm=−40 cm=−30 cm (PI1=40 cm) Using the mirror formula, f2=−30 cm v1+u1v21+−401v2=f1=−301=P2I2=−120 cm

Q: Focal length of the concave mirror shown in figure is 20 cm. and ab=1 mmbc=2 mm For the given situation, make its image

u=−30 cm and f=−20 cm Using the mirror formula, v1+u1=f1, we have v1+−301=−201 Solving this equation, we get Now, v=−60 cmm=−uv=−(−30)(−60)=−2 Now c point is 2 mm above the principal axis and magnification is -2 . Hence, image c′ will be formed 4 mm below the principal axis. Similarly a point is 1 mm below the principal axis and value of m is -2 . Hence, image a′ is formed 2 mm above the principal axis. Image with ray diagram of c is as shown below.

Q: Draw u versus v graph or u1 versus v1 graph for a concave mirror of focal length f.

The mirror formula is v1+u1=f1 If we take v1 along y-axis and u1 along x-axis, then the above equation becomes y+x=c( as f1= constant ) Therefore, v1 versus u1 graph will be a straight line. Let us take origin at pole. u versus v and u1 versus v1 graphs are as shown below.

Question 1

Find the distance of object from a concave mirror of focal length 10 cm so that image size is four times the size of the object.

Accepted Answer

Concave mirror can form real as well as virtual image. Here, nature of image is not given in the question. So, we will consider two possible cases.
Case 1 (When image is real) Real image is formed on the same side of the object, i.e. u,v and f all are negative. So let,
then
and
Substituting in
We have
∴
u=−x
v=−4x as ∣∣​uv​∣∣​=∣m∣=4f=−10 cmv1​+u1​=f1​​
−4x1​−x1​x​=−101​ or 4x5​=101​=12.5 cm​
Case 2 (When image is virtual) In case of a mirror, image is virtual, when it is formed behind the mirror, i.e. u and f are negative, while v is positive. So let,
u=−y
then
v=+4y and f=−10 cm
Substituting in
v1​+u1​4y1​−y1​4y3​y​=f1​=−101​=101​=7.5 cm​
or

Question 2

Find the condition under which a convex mirror can make a real image.

Accepted Answer

Substituting the sign of f only in the mirror formula, we have
v1​+u1​=+f1​⇒v1​=f1​−u1​
For real image v should be negative and for this u should be positive and less than f. Object distance u is positive means object should be virtual and lying between P and F. The ray diagram is as shown below.

Here, O= virtual between P and F I= real

Question 3

Focal length of the mirror shown in figure is 20 cm. Find the image position and its velocity.

Accepted Answer

Substituting the values,
u=−30 cm and f=−20 cm
in the mirror formula, we have
v1​+−301​=−201​
Solving this equation, we get
Further,
v=−60 cm
m=−uv​=−(−30)(−60)​=−2
and
m2=4
Object velocity along the axis is 5cos37∘=4 mm/s (towards OP ). Therefore, image velocity along the axis should be m2 times or 16 mm/s in the opposite direction of object velocity. Object velocity perpendicular to axis is 5sin37∘=3 mm/s (upwards). Therefore, image velocity will be m times or −6 mm/s downwards. The position and velocity of image is shown below.

vI​tanθ​=(16)2+(6)2​=292​ mm/s=166​=83​ or θ=tan−1(83​)​

Question 4

A square mnqp is kept between F and C on the principal axis of a concave mirror as shown in figure. Find a rough image of this object.

Accepted Answer

Object is placed between F and C, therefore, image is real, inverted magnified and beyond C. Further pq is towards F. Therefore, its image p′q′ is towards infinity and it is more magnified. A rough image is as shown alongside.

From the ray shown in figure, we can see that n′q′ will be a straight line.
Exercise In the given problem, focal length of mirror is 30 cm and side of square is 10 cm with pP=40 cm. Find perimeter of the image
Ans (90+1510​)cm

Question 5

Focal length of convex mirror M1​ is 20 cm and that of concave mirror M2​ is 30 cm. Find position of second image I2​. Take first reflection from M1​.

Accepted Answer

For M1​
Using the mirror formula,
u1​=−20 cmf1​=+20 cm​
Solving, we get
For M2​
v1​+u1​=f1​, we have
v1​1​+−201​v1​u2​f2​​=+201​=+10 cm=−40 cm=−30 cm​
(PI1​=40 cm)
Using the mirror formula,
f2​=−30 cm
v1​+u1​v2​1​+−401​v2​​=f1​=−301​=P2​I2​=−120 cm​

Question 6

Focal length of the concave mirror shown in figure is 20 cm.
and
ab=1 mmbc=2 mm​
For the given situation, make its image

Accepted Answer

u=−30 cm and f=−20 cm
Using the mirror formula, v1​+u1​=f1​, we have
v1​+−301​=−201​
Solving this equation, we get
Now,
v=−60 cmm=−uv​=−(−30)(−60)​=−2​
Now c point is 2 mm above the principal axis and magnification is -2 . Hence, image c′ will be formed 4 mm below the principal axis. Similarly a point is 1 mm below the principal axis and value of m is -2 . Hence, image a′ is formed 2 mm above the principal axis. Image with ray diagram of c is as shown below.

Question 7

A small object ab of size 1 mm is kept at a distance of 40 cm from a concave mirror of focal length 30 cm. Make image of this object.

Accepted Answer

Here, object size is very small. So, this is case 1 .
and
u=−40 cmf=−30 cm​
Applying the mirror formula
v1​+u1​v1​+−401​​=f1​, we have =−301​​
Solving this equation, we get
Further,
vm=−uv​m2LI​=m2Lo​​=−120 cm=−(−40)(−120)​=−3=9=(9)(1 mm)=9 mm​
Now,
Image diagram is as shown below.

Question 8

A thin rod of length 3f​ is lying along the principal axis of a concave mirror of focal length f. Image is real, magnified and inverted and one of the end of rod coincides with its image itself. Find length of the image.

Accepted Answer

Image is real, magnified and inverted. So, the given rod lies between F and C. Further, one end of the rod is coinciding with its image itself. Therefore, it is lying at C. So, the thin rodCR is kept as shown below.

We have to apply mirror formula only for point R.
u=−35f​, focal length =−fv1​+u1​=f1​,v1​+−35f​1​=−f1​​
Using the mirror formula
We have,
v1​+−35f​1​=−f1​
Solving this equation, we get
v=2−5f​ or −2.5f
So, the image of rod CR is C′R′ as shown below.

Question 9

Draw u versus v graph or u1​ versus v1​ graph for a concave mirror of focal length f.

Accepted Answer

The mirror formula is
v1​+u1​=f1​
If we take v1​ along y-axis and u1​ along x-axis, then the above equation becomes
y+x=c( as f1​= constant )
Therefore, v1​ versus u1​ graph will be a straight line. Let us take origin at pole.

u versus v and u1​ versus v1​ graphs are as shown below.

Question 10

An object is 30.0 cm from a spherical mirror, along the central axis. The absolute value of lateral magnification is 21​. The image produced is inverted. What is the focal length of the mirror?

Accepted Answer

Image is inverted, so it is real. u and v both are negative. Magnification is 21​, therefore, v=2u​.
Given, u=−30 cm,v=−15 cm
Using the mirror formula,
v1​+u1​=f1​
 We have, f1​=−151​−301​=−101​
∴f=−10 cm

Question 11

A concave mirror has a radius of curvature of 24 cm. How far is an object from the mirror if an image is formed that is :
(a) virtual and 3.0 times the size of the object,
(b) real and 3.0 times the size of the object and
(c) real and 1/3 the size of the object?

Accepted Answer

Given, R=−24 cm
(concave mirror)
Hence, f=2R​=−12 cm
(a) Image is virtual and 3 times larger. Hence, u is negative and v is positive. Simultaneously, ∣v∣=3∣u∣. So let,
u=−x
v=+3x
then
Substituting in the mirror formula,
v1​+u1​=f1​
We have,
3x1​−x1​=−121​
x=8 cm
Therefore, object distance is 8 cm.
(b) Image is real and three times larger. Hence, u and v both are negative and ∣v∣=3∣u∣.
So let,
u=−x
then
Substituting in mirror formula, we have
−3x1​−x1​=−121​
or
x=16 cm
∴ Object distance should be 16 cm.
(c) Image is real and 31​ rd the size of object. Hence, both u and v are negative and ∣v∣=3∣u∣​.
So let,
u=−x
v=−3x​
then
Substituting in the mirror formula, we have
−x3​−x1​=−121​
∴x=48 cm
∴ Object distance should be 48 cm.

Question 12

A ray of light is incident on a plane mirror along a vector i^+j^​−k^. The normal on incidence point is along i^+j^​. Find a unit vector along the reflected ray.

Accepted Answer

Reflection of a ray of light is just like an elastic collision of a ball with a horizontal ground. Component of incident ray along the inside normal gets reversed while the component perpendicular to it remains unchanged. Thus, the component of incident ray vector A=i^+j^​−k^ parallel to normal, i.e. i^+j^​ gets reversed while perpendicular to it, i.e. −k^ remains unchanged. Thus, the reflected ray can be written as
R=−i^−j^​−k^
∴ A unit vector along the reflected ray will be
r^=RR​=3​−i^−j^​−k^​
r^=−3​1​(i^+j^​+k^)

Question 13

A gun of mass m1​ fires a bullet of mass m2​ with a horizontal speed v0​. The gun is fitted with a concave mirror of focal length f facing towards a receding bullet. Find the speed of separations of the bullet and the image just after the gun was fired.

Accepted Answer

Let v1​ be the speed of gun (or mirror) just after the firing of bullet. From conservation of linear momentum,
m2​v0​=m1​v1​
v1​=mm2​v0​​
or

Now, dtdu​ is the rate at which distance between mirror and bullet is increasing =v1​+v0​
We have already read in extra points that :
∴
dtdv​=(u2v2​)dtdu​
Here,
u2v2​=m2=1 (as at the time of firing, bullet is at pole).
dtdv​=dtdu​=v1​+v0​

Here, dtdv​ is the rate at which distance between image (of bullet) and mirror is increasing. So, if v2​ is the absolute velocity of image (towards right), then
v2​−v1​v2​​=dtdv​=v1​+v0​=2v1​+v0​​
Therefore, speed of separation of bullet and image will be
vr​vr​​=v2​+v0​=2v1​+v0​+v0​=2(v1​+v0​)​
Substituting value of v1​ from Eq. (i), we have
vr​=2(1+m1​m2​​)v0​

Question 14

Assertion : A convex mirror can never make a real image. Reason : For all real objects image formed by a convex mirror is virtual.

Accepted Answer

Convex mirror can make real images of virtual objects.

Question 15

Assertion : Real view mirror of vehicles is a convex mirror. Reason : It never makes real image of real objects.

Accepted Answer

Field of view of convex mirror is large.

Question 16

Assertion : If magnification of a real object is −2. Then, it is definitely a concave mirror. Reason : Only concave mirror can make real images of real objects.

Accepted Answer

m=−2, means image is real, inverted an 2-times magnified.

Question 17

Assertion: Any ray of light suffers a deviation of (180∘−2i) after one reflection. Reason : For normal incidence of light deviation is zero.

Accepted Answer

For normal incidence, i=0∘∴δ=180∘

Question 18

Assertion : In case of a concave mirror if a point object is moving towards the mirror along its principal axis, then its image will move away from the mirror. Reason: In case of reflection (along the principal axis of mirror) object and image always travel in opposite directions.

Accepted Answer

(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

Question 19

Assertion : Two plane mirrors kept at right angles deviate any ray of light by 180∘ after two reflections. Reason : The above condition is satisfied only for angle of incidence i=45∘.

Accepted Answer

If Assertion is true, but the Reason is false.

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Solutions for Optics And Modern Physics by DC Pandey

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