Tag: reflection of light in spherical mirrors

Questions Related to reflection of light in spherical mirrors

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


Correct Option: C

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. $200\, cm, -100$


Correct Option: A

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. $\pm 6.8\ mm$


Correct Option: A

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


Correct Option: A
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

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.


Correct Option: A

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. $97 cm$


Correct Option: B

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


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 $f = 20 \,cm$.

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

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

  3. $2$

  4. $4$


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:

  1. $(n-1)f$

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

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

  4. $(n+1)f$


Correct Option: C

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


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.