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Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A convex lens of focal length 0.12 m produces a virtual n image which is thrice the size of the object. Find the distance between the object and the lens

  1. 0.04 m

  2. 0.08 m

  3. 0.12 m

  4. 0.24 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From lens formula
$\dfrac{{1}{v}-\dfrac{1}u}=\dfrac{1}{f}$
$\dfrac{u}{v}-1=\dfrac{u}{f}$
$\dfrac{u}{v}=\dfrac{u+f}{f}$
$\dfrac{v}{u}=\dfrac{f}{u+f}$
For a convex lens f = 0.12m and m=3 then find u
So ,  $3=\dfrac{0.12}{0.12+u}$
$0.36+3u = 0.12$
$u=-0.08 m$

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

a) For concave mirror focal length is taken as ........
b) For convex mirror, radius of curvature is taken as ........

  1. $a = positive, b = negative$
  2. $a = negative, b = positive$
  3. $a = positive, b = positive$
  4. $a = negative, b = negative$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to cartesian system focal length of a concave mirror is negative whereas radius of curvature of convex mirror is positive.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A $2.0 cm$ object is placed $15 cm$ in front of a concave mirror of focal length $10 cm$. What is the size and nature of the image?

  1. $4 cm$, real
  2. $4 cm$, virtual
  3. $1.0 cm$, real

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac { 1 }{ v } -\dfrac { 1 }{ 15 } =\dfrac { 1 }{ -10 } $
$\Rightarrow v=-30cm$
$\therefore m=-\dfrac { v }{ u } =-\dfrac { -30 }{ 15 } =2$
So, image size will be $4 cm$ (real)

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A $2.0\ cm$ long object is placed perpendicular to the principal axis of a concave mirror. The distance of the object from the mirror is $30\ cm$, and its image is formed $60\ cm$ from the mirror, on the same side of the mirror as the object. Find the height of the image formed :

  1. $3\ cm$
  2. $4\ cm$
  3. $5\ cm$
  4. $8\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Distance of object from the mirror, $u=-30$ 


Distance of image from the mirror, $v=-60$ , image is real.


Now we know  magnification ratio $m=-\dfrac{v}{u}$

$m=-\dfrac{-60}{-30}$

$m=-2$

Height of object $h=2 cm$

Image height $h _1=m\times h=-2\times 2=-4cm$

Height of image is 4 cm and its inverted .

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

Magnification m = _____

  1. v/u

  2. u/v
  3. $h _0/h _i$
  4. $h _i/h _0$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

The magnification equation relates the ratio of the image distance(v) and object distance(u) and also to the ratio of the image height ($h _i$) and object height ($h _o$).

i.e., $m=\dfrac vu=\dfrac {h _i}{h _o}$

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

In a car a rear view mirror having a radius of curvature 1.50 m forms a virtual image of a bus located 10.0 m from the mirror. The factor by which the mirror magnifies the size of the bus is close to

  1. 0.06

  2. 0.07

  3. 0.08

  4. 0.09
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

R=1.5, u=$-10$

$\dfrac{2}{R}=\dfrac{1}{v}+\dfrac{1}{u}$
$v=\dfrac{30}{43}$
$m=\dfrac{-v}{u}=.069=.07$

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A convex lens is used to form an image of an object on a screen. If the upper half of the lens is blackened so that it becomes opaque, then

  1. Only half of the image will be visible

  2. The image position shifts towards the lens

  3. The image position shifts away from the lens

  4. The brightness of the image reduces
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

To form a image only two rays are needed .The total amount of light released by the object is not allowed to pass through the lens, intensity of image will decrease

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

An object is kept on the principal axis of a concave mirror of focal length $10$cm, at a distance of $15$cm from its pole. The image formed by the mirror is?

  1. Virtual and magnified

  2. Virtual and diminished

  3. Real and magnified

  4. Real and diminished
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Focal length of concave mirror  $f = -10 \ cm$
Object distance   $u = -15 \ cm$
Using     $\dfrac{1}{v}+\dfrac{1}{u} = \dfrac{1}{f}$
$\therefore$   $\dfrac{1}{v}+\dfrac{1}{-15} = \dfrac{1}{-10}$
$\implies   \ v = -30 \ cm$
Since $v$ is negative, so image formed is real.
Magnification  $m = \dfrac{-v}{u}$
$\therefore$  $m = -\dfrac{(-30)}{-15} = -2$
Since $m$ is greater than $1$, so magnified image is formed.
Correct answer is option C.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

From the understanding of cartesian sign convention for reflection by spherical mirror,  students took part in group discussion for FA - 1 in classroom. Who is wrong in the group discussion?
Alpesh : All the distances are measured from the pole of a mirror parallel to the principal axis.
Beena : The distances measured in the direction of incident ray are taken positive.
Chamak : The height measured upward and perpendicular to principal axis is taken negative.
Daksha : The height measured downward and perpendicular to principal axis is taken as positive.

  1. Champak and Daksha

  2. Only Champak

  3. Alpesh and Beena

  4. Only Daksha
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

1- It is true that all the distances are measured from the pole of a mirror parallel to the principal axis.So, Alpesh is right.

2-It is also true that the distances measured in the direction of incident ray are taken positive.So, Beena is also right
3-The height measured upward and perpendicular to principal axis is taken positive. So, Chamak is wrong.
4-The height measured downward and perpendicular to principal axis is taken as negative.So, Daksha is also wrong.
So, only Champak and Daksha are wrong.
Therefore, A is correct option.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

Photographs of the ground are taken from an aircraft flying at an altitude of 2000 m by a camera with a lens of focal length 50 cm. The size of the film in the camera is $18 cm\times 18 cm.$ The area of the ground that can be photographed by the camera is:

  1. $720 m\times 720 m$
  2. $720 cm\times 720 cm$
  3. $360 cm\times 360 cm$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac {\text {Area of the image}}{\text {Area of the object}}=$ Aerial magnification $=m^2$
Where m is the linear magnification
$\therefore m=\dfrac {f}{f+u}=\dfrac {0.5}{0.5 - 2000} \approx -\dfrac {1}{4000}$
Now, area of the image is equal to size of the camera film, therefore
Area of object$=\dfrac {\text {Area of the image}}{m^2}$
$=18\times 18\times 4000\times 4000$
$=72000 cm\times 72000 cm$
Note that m has no unit, so area of object is in same units as area of image.