Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A convex lens is given, for which the minimum distance between an object and its rel image is $40cm$. An object is placed at a distance of $15cm$ from this lens. The liner magnification of adjustment will be 

  1. $\dfrac{5}{3}$
  2. $-2$
  3. $2$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given object distance $u=15$ cm

Distance between object and real image produced $=40 $cm
Thus image distance $v=40-15=25$ cm
Also we know linear magnification,
$m=\dfrac{-v}{u}=\dfrac{-25}{-15}=\dfrac{5}{3}$ 

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

An object is placed in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm, the position and nature of the image formed is :

  1. $+30 cm, virtual \ and \ erect$
  2. $+30 cm, real \ and \ inverted$
  3. $-30 cm, virtual \ and \ erect$
  4. $-30 cm, real \ and \ inverted$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

An object of length $6\ cm$ is placed on the principle axis of a concave mirror of focal length $f$ at a distance of $4\ f$. The length of the image will be

  1. $2\ cm$
  2. $12\ cm$
  3. $4\ cm$
  4. $1.2\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

The object distance $u=-6\,cm$

Now, magnification is

  $ m=\dfrac{I}{O} $

 $ m=\dfrac{f}{f-u} $

 $ \dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)} $

 $ I=-2\,cm $

Hence, the length of image is -$2\ cm$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

An astronomical telescope has focal lengths $100$ & $10$cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision,magnification power of telescope will be,

  1. -15

  2. -14

  3. -17

  4. -19
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given focal length of eye piece${f} _{e}=10cm\$

focal length of objective${f} _{o}=100 cm\$
Also we know least distance $D=25 cm\$ 
Magnifying power $M=\dfrac{-{f} _{0}}{{f} _{e}}(1+\dfrac{{f} _{e}}{D})\$
$M=-\dfrac{100}{10}(1+\dfrac{10}{25})\$
$M=-14$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

In the displacement method, a convex lens is placed in between an object and a screen. If one of the magnification is $3$ and the displacement of the lens between the two positions is $24$cm, then the focal length of the lens is:

  1. $10$ cm
  2. $9$ cm
  3. $6$ cm
  4. $16/3$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given magnification $M=\dfrac{v}{u}=3$


Thus $v=3u$, where v and u are the image and object distance respectively.

Also Distance between lenses$=v-u=24$
Thus $u=12 cm$, than $v=36 cm$

From lens formula we have,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{f}=\dfrac{1}{36}+\dfrac{1}{12}$

$\dfrac{1}{f}=\dfrac{4}{36}$

$f=\dfrac{36}{4}$

$f=9 cm$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A concave mirror of focal length $20\ cm$ produces an image twice the height of the object. If the image is real, then the distance of the object from the mirror is:

  1. $20\ cm$
  2. $60\ cm$
  3. $10\ cm$
  4. $30\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a real image, magnification m = -v/u = -2. Thus v = 2u. Using 1/v + 1/u = 1/f with f = -20 cm, we have 1/(2u) + 1/u = 1/-20. This simplifies to 3/(2u) = -1/20, so 2u = -60, u = -30 cm. The distance is 30 cm.

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

In a concave mirror an object is placed at a distance x from the focus, and the image is formed at a distance y from the focus. The focal length of the mirror is

  1. $xy$
  2. $\sqrt{xy} $
  3. $\dfrac{x+y}{2} $
  4. $\sqrt{\dfrac{x}{y} }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a property of concave mirrors where the focal length f is the geometric mean of the distances of the object and image from the focus. Specifically, f^2 = x * y, so f = sqrt(xy).

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A light ray travelling parallel to the principle axis of a concave mirror strikes the minor at angle of incidence $\theta$. If the radius of curvature of the mirror is $R$, then after reflection, the ray meets the principle axis at distance $d$ from the centre of curvature, then $d$ is 

  1. $\dfrac {R}{2}$
  2. $R\left(1-\dfrac {1}{2\cos \theta}\right)$
  3. $\dfrac {R}{2\cos \theta}$
  4. $\dfrac {R}{2}(1+\cos \theta)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a ray parallel to the axis striking at angle of incidence theta, the distance from the pole to the intersection point is f/cos(theta) = (R/2)/cos(theta). The distance from the center of curvature (at R) is R - (R/2)/cos(theta). None of the options match this standard derivation perfectly, but R/2 is the paraxial approximation.

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

The focal length of a concave mirror is f and the distance from the object to the principal focus is p. The ratio of the size of the real image to the size of the object is:

  1. $-\displaystyle \frac{f}{p}$
  2. $\displaystyle \left(\frac{f}{p}\right)^2$
  3. $\displaystyle \left(\frac{f}{p}\right)^{\frac{1}{2}}$
  4. $-\displaystyle \frac{p}{f}$
  5. $-fp$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance of object is $u= -(f+p)$
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ gives:
$\dfrac{1}{v}-\dfrac{1}{f+p}= -\dfrac{1}{f}$
or, $\dfrac{1}{v}= -\dfrac{1}{f}+\dfrac{1}{f+p}$
or, $\dfrac{1}{v}= -\dfrac{p}{(f+p)\times f}$
or, $v= -\dfrac{(f+p)\times f}{p}$      (-ve sign indicates image is real) 
   Magnification $=-\dfrac{v}{u}$ 
           $=-\dfrac{(f+p)\times f}{p\times (f+p)}$   
           $=-\dfrac{f}{p}$  (-ve sign indicates inverted)
    So, ratio of size of image to that of object is: $-\dfrac{f}{p}$