Tag: sign convention for the measurement of distances

Questions Related to sign convention for the measurement of distances

A point object is placed at a distance of  $10\mathrm { cm }$  and its real image is formed at a distance of  $30\mathrm { cm }$  from a concave mirror. If the object is moved by  $0.2\mathrm { cm }$  towards the mirror. the image will shift by about.

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

  1. $1.8\mathrm { cm }$ away from the mirror

  2. $0.4\mathrm { cm }$ towards the mirror

  3. $0.8\mathrm { cm }$ away from the mirror

  4. $0.8\mathrm { cm }$ towards the mirror

Show answer
Correct Option: B

For position of real object at $x _1$ and $x _2 (x _2 > x _1)$ magnification is equal to $2$. Find out $\dfrac{x _1}{x _2}$. if focal length of converging lens $f = 20 \,cm$.

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

  1. $\dfrac{1}{2}$

  2. $\dfrac{1}{4}$

  3. $2$

  4. $4$

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Correct Option: A
Explanation:

$m = \left(\dfrac{f}{f + u}\right)$

$-2 = \dfrac{20}{20 - x _2}$

$-10 x _2 = 10$
$x _2 = 20 \,cm$

$m = 2 = \dfrac{20}{20 - x _1}$

$20 - x _1 = 10$
$x _1 = 10$

$\dfrac{x _1}{x _2} = \dfrac{10}{20} = \left(\dfrac{1}{2}\right)$

A concave mirror produces an image n times the size of an object. If the focal length of the mirrors is '$f$' and image formed is real, then the distance of the object from the mirror is:

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

  1. $(n-1)f$

  2. $\dfrac{(n-1)}{n}f$

  3. $\dfrac{(n+1)}{n}f$

  4. $(n+1)f$

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Correct Option: C

Which one of the following has a negative sign, on the basis of new Cartesian sign Convention?

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

  1. Image distance for a convex mirror

  2. Height of a virtual and erect image

  3. Focal length of a convex mirror

  4. Object distance for a concave mirror

Show answer
Correct Option: D
Explanation:

According to new Cartesian sign convention, object distance for any lens or mirror is measured as negative. This is because, the object distance is measured against the direction of incident light.

So, for the given options, object distance for a concave mirror has a negative sign.

Which of the following statements is correct according to New Cartesian Sign Convention?

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

  1. Focal length of a concave mirror is positive and that of a convex mirror is negative.

  2. Focal length of a concave mirror is negative and that of a convex mirror is positive.

  3. Image distance is always positive for a concave mirror.

  4. The height of all the real and inverted images is positive.

Show answer
Correct Option: B
Explanation:

The focal length of a concave mirror, with a real focus, is always positive as is measured against the direction of incident light.

The focal length of a convex mirror, with a virtual focus, is always negative as is measured along the direction of the incident light.

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

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

  1. 0.04 m

  2. 0.08 m

  3. 0.12 m

  4. 0.24 m

Show answer
Correct Option: B
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 ,  $3=\dfrac{0.12}{0.12+u}$
$0.36+3u = 0.12$
$u=-0.08 m$

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

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

  1. $a = positive, b = negative$

  2. $a = negative, b = positive$

  3. $a = positive, b = positive$

  4. $a = negative, b = negative$

Show answer
Correct Option: B
Explanation:

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

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?

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

  1. $4 cm$, real

  2. $4 cm$, virtual

  3. $1.0 cm$, real

  4. None of these

Show answer
Correct Option: A
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)

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 :

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

  1. $3\ cm$

  2. $4\ cm$

  3. $5\ cm$

  4. $8\ cm$

Show answer
Correct Option: B
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 .

Magnification m = _____

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

  1. v/u

  2. u/v

  3. $h _0/h _i$

  4. $h _i/h _0$

Show answer
Correct Option: A,D
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., $m=\dfrac vu=\dfrac {h _i}{h _o}$