Tag: reflection of light in spherical mirrors

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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 thin. rod of length f/ 3 is placed along the principal axis of a concave mirror of focal length f such that its image which is real and elongated, just touches one end of the rod. What is its magnification ?

  1. +2

  2. -3

  3. -1.5

  4. -2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The rod of length f/3 is placed along the axis. Let the ends be at u1 and u2. The image ends are at v1 and v2. Magnification m = dv/du. Using 1/v + 1/u = 1/f, differentiating gives dv/v^2 = -du/u^2, so m = -v^2/u^2. Given the geometry, the calculation leads to m = -1.5.

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 astronomical telescope has an objective of focal length $200 \,cm$ and an eye piece of focal length $4\,cm$ The telescope is focused to see an object $10\, km$ from the objective,.The final image is formed at infinity. The length of the tube and angular magnification produced by it is

  1. $204\, cm, -50$
  2. $200\, cm, -50$
  3. $204\, cm, -100$
  4. $200\, cm, -100$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an astronomical telescope, the length of the tube is fo + fe = 200 + 4 = 204 cm. The angular magnification for a distant object is -fo/fe = -200/4 = -50.

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

Let the equation connecting object distance $u$, image distance $v$ and focal length $f$ for a lens be $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$. A student measures values of $u$ and $v$, with their associated uncertainties.
These are $u = 50\ mm \pm 3\ mm, v = 200\ mm \pm 5\ mm$. He calculates the value of $f$ as $40\ mm$. What is the uncertainty in this value?

  1. $\pm 2.1\ mm$
  2. $\pm 3.4\ mm$
  3. $\pm 4.5\ mm$
  4. $\pm 6.8\ mm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 1/f = 1/u + 1/v, the uncertainty is df/f^2 = du/u^2 + dv/v^2. Plugging in the values: df = f^2 * (du/u^2 + dv/v^2) = 40^2 * (3/50^2 + 5/200^2) = 1600 * (3/2500 + 5/40000) = 1600 * (0.0012 + 0.000125) = 1600 * 0.001325 = 2.12 mm.

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 man has a concave shaving mirror of focal length $0.2$ m. How far should the mirror be held from his face in order to give an image of two fold magnification? 

  1. $0.1$ m
  2. $0.2$ m
  3. $0.3$ m
  4. $0.4$ m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Concave shaving mirror  $f = 0.2m$

Magnification,$m = 2$
$m = \dfrac{-v}{u}$

$2 = \dfrac{-v}{u}$
$v = -2u$
Using mirror formula 
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$- \dfrac{1}{0.2} = -\dfrac{1}{2u} + \dfrac{1}{u}$

$-\dfrac{1}{0.2}=\dfrac{1}{2u}$

$u = -0.1m$
i.e shaving mirror should be 10cm ahead of man

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 small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm.

  1. The distance from the mirror, should a screen be placed in order to receive a sharp image is-54 cm.

  2. The nature of image is virtual inverted w.r.t. object.

  3. The image formed is 8 times highest the object.

  4. The image formed is 3 times highest the object.
Reveal answer Fill a bubble to check yourself
A Correct answer
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 placed at a distance of 36 cm from a convex mirror . A plane mirror is placed in between , so that the two virtual image so formed coincide . If the plane mirror is at a distance of 24 cm from the object , find the radius of curvature of the convex mirror . 

  1. $43 cm$
  2. $36 cm$
  3. $78 cm$
  4. $97 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 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.

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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 $f = 20 \,cm$.

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

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

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

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

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

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

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

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.