Tag: reflection of light at curved surfaces

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Questions Related to reflection of light at curved surfaces

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

Sun subtends an angle of $0.5^{o}$ at the pole of a concave mirror of radius of curvature 15 m. The diameter of the image of the sun formed by the mirror is

  1. $8.55 cm$
  2. $7.55 cm$
  3. $6.55 cm$
  4. $5.55 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The radius of curvature R is 15 m, so the focal length f = R/2 = 7.5 m. The sun subtends an angle theta = 0.5 degrees = 0.5 * (pi/180) radians. The diameter of the image d is given by f * theta, which evaluates to (7.5 m) * (0.5 * pi / 180) approximately equal to 0.0654 m or 6.55 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 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: $-\dfrac{f}{p}$
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 glass hemisphere of radius R and of material having refractive index 1.5 is silvered on its flat face as shown in figure . a small object of height h is located at distance 2R from the surface of hemisphere as shown in the figure. the final image will form

  1. At a distance of R from silvered surface, on the right side

  2. on the object itself

  3. at hemisphere surface

  4. refractive index
Reveal answer Fill a bubble to check yourself
B 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

A small piece of wire bent into L shape such that the upright and horizontal portions are of equal length. It is placed with the horizontal portion along the axis of concave mirror of radius of curvature 20 cm. If the bend is 40 cm from the pole of the mirror, then the ratio of the length of the images of the upright and horizontal portions of the wire is

  1. 1 : 9

  2. 1 : 3

  3. 3 : 1

  4. 2 :1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$u= -40$ ; $f= -10$

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

$\dfrac{1}{v}-\dfrac{1}{40}=\dfrac{-1}{10}$

$v=\dfrac{-40}{3}$

magnification=$\dfrac{-v}{u}=\dfrac{-1}{3}$

lateral magnification is $\dfrac{f^{2}}{(u-f)^{2}}$

                                 =  $\dfrac{100}{(40-10)^{2}}$

                                 =  $\dfrac{1}{9}$

ratio of up-right portion to lateral portion is $\dfrac{\dfrac{1}{3}}{\dfrac{1}{9}}=3:1$, hence option $C$ is correct 
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 small piece of wire bent into an L shape, with upright and horizontal portions of equal lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is 10 cm. If the bend is 20 cm from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of the wire is :

  1. 1:2

  2. 3:1

  3. 1:3

  4. 2:1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, $u= -20$ ; $f= -5$

From mirror formula, $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{v}-\dfrac{1}{20}=\dfrac{-1}{5}$

$v=\dfrac{-20}{3}$

magnification=$\dfrac{-v}{u}=\dfrac{-1}{3}$

lateral magnification is $\dfrac{f^{2}}{(u-f)^{2}}$

                                 =  $\dfrac{25}{(20-5)^{2}}$

                                 =  $\dfrac{1}{9}$
Ratio of up-right portion to lateral portion is $\dfrac{\dfrac{1}{3}}{\dfrac{1}{9}}=3:1$

option $B$ is correct