Tag: sign convention for mirrors

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Questions Related to sign convention for mirrors

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

Linear magnification is

  1. Positive for an inverted image

  2. Negative for an erect image

  3. Zero for an inverted image

  4. Negative for an inverted image
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Linear magnification is given as Image size/ Object size

For calculations of magnification, values of image size and object size are taken with proper sign convention. 
For an inverted image, the image size is taken as negative and hence magnification is negative.

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

Distances measured below the principal axis are taken as ........

  1. Positive

  2. Negative

  3. None

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

According to the sign convention followed for mirrors and for lenses, the distances measured above the principal axis are taken as positive and the distances measured below the principal axis are taken as negative.

Hence, the correct answer is OPTION B.

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

State whether true or false : 

The magnification produced by a spherical mirror is the ratio of the height of the image to the height of the object.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The magnification is the ratio of height of the image to height of the object. The image is smaller than the object, the magnification will be less than 1. If the image if upside down (inverted) then magnification will be negative.

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 concave mirror forms an erect image of an object placed at a distance of 10 cm from it. The size of the image is double that of the object. Where is the image formed?

  1. 20 cm behind the mirror

  2. 20 cm in front of the mirror

  3. 40 cm behind the mirror

  4. 40 cm in front of the mirror
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the image is erect , so the magnification ratio must be positive .


$m=2$

also $u=-10cm$

so $m=-\dfrac{v}{u}$

$2=-\dfrac{v}{-10}$ 

$v=20cm$

So the image is virtual and 20 cm from mirror on the other side of the object.

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 dentist wants a small mirror that when placed $2$cm from a tooth, will produce $3\times$ upright image. What kind of mirror must be used and what must its focal length be?

  1. Concave mirror, $3.04$ cm
  2. Concave mirror, $1.5$ cm
  3. Convex mirror, $3.0$ cm
  4. Convex mirror, $1.5$cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$m=+\dfrac{v}{u}=\dfrac{+f}{u-f}$
$3=\dfrac{+f}{+2+f}\Rightarrow f=3cm$.

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

The focal length of a concave mirror is 50 cm where an object is to be placed so that its image is two times magnified, real and inverted :

  1. 75 cm

  2. 72 cm

  3. 63 cm

  4. 50 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

Focal length, $f=50\,cm$

Magnification, $m=2$

$ m=\dfrac{v}{u} = 2$

$ v=2u $

From mirror formula,

$ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} $

$ \dfrac{1}{f}=\dfrac{1}{2u}+\dfrac{1}{u}$

$ u=\dfrac{3f}{2}=\dfrac{3\times 50}{2}=75\ cm $

Object Is placed at $75\,cm$ from the mirror.

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 mirror has a focal length $f$.A real object is placed at a distance $f$ in front of it from the pole, produces an image at:

  1. $\infty$
  2. $f$
  3. $\dfrac{f}{2}$
  4. $2f$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a convex mirror, f is positive. With object distance u = -f, the mirror formula 1/v + 1/u = 1/f becomes 1/v - 1/f = 1/f. Thus, 1/v = 2/f, so v = f/2.

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

The distance between an object its doubly magnified by a concave mirror of focal length $f$ is

  1. $3 f/2$
  2. $2 f/3$
  3. $3\ f$

  4. Depend on whether the image is real or virtual
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Focal length $=f$

We know that, the magnification is

$m=\dfrac{-v}{u}$

Now, magnification is double and image is real

  $ -2=\dfrac{-v}{u} $

 $ v=2u $

Now, using formula of mirror

  $ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} $

 $ \dfrac{1}{f}=\dfrac{1}{2u}+\dfrac{1}{u} $

 $ \dfrac{1}{f}=\dfrac{3}{2u} $

 $ u=\dfrac{3f}{2} $

Hence, the distance is $\dfrac{3f}{2}$ 

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 $2.5\ cm$ high is placed at a distance of $10\ cm$ from a concave mirror of radius of curvature $30\ cm$. The size of the image is:

  1. $9.2\ cm$
  2. $10.5\ cm$
  3. $5.6\ cm$
  4. $7.5\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,

Distance $u=-10\,cm$

Height $h=2.5\,cm$

Radius of curvature $R=30\,cm$

We know that,

  $ -f=\dfrac{R}{2} $

 $ -f=\dfrac{30}{2} $

 $ f=-15\,cm $

Now, using mirror’s formula

  $ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} $

 $ -\dfrac{1}{15}=\dfrac{1}{v}-\dfrac{1}{10} $

 $ \dfrac{1}{v}=-\dfrac{1}{15}+\dfrac{1}{10} $

 $ v=30\,cm $

Now, the magnification is

  $ m=\dfrac{-v}{u} $

 $ m=\dfrac{30}{10} $

 $ m=3 $

We know that,

  $ m=\dfrac{h'}{h} $

 $ 3=\dfrac{h'}{2.5} $

 $ h'=7.5\,cm $

Hence, the size of the image is $7.5\ cm$

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

The focal length $f$ of a mirror is given by $\cfrac{1}{f}=\cfrac{1}{u}+\cfrac{1}{v}$, where $u$ and $v$ represent object and image distances, respectively

  1. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}$
  2. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{v}+\cfrac{\Delta v}{u}$
  3. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}-\cfrac{\Delta (u+v)}{u+v}$
  4. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}+\cfrac{\Delta U}{u+v}+\cfrac{\Delta v}{u+v}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The focal length $f$ of a mirror is given by
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$. . . . . . . .(1)
where $v=$ image distance
$u=$ object distance
from equation (1),
$f=\dfrac{uv}{u+v}=uv(u+v)^{-1}$
Taking log both sides we get
$logf=logu+log v+log(u+v)^{-1}$
$log f=logu+logv-log(u+v)$
Differentiating with respect to each variable
$\dfrac{\Delta f}{f}=\dfrac{\Delta u}{u}+\dfrac{\Delta v}{v}+\dfrac{\Delta (u+v)}{u+v}$
$\dfrac{\Delta f}{f}=\dfrac{\Delta u}{u}+\dfrac{\Delta v}{v}+\dfrac{\Delta u}{u+v}+\dfrac{\Delta v}{u+v}$
The correct option is D.