Tag: reflection of light at curved surfaces

Questions Related to reflection of light at curved surfaces

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:

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

  1. $10$ cm

  2. $9$ cm

  3. $6$ cm

  4. $16/3$ cm

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Correct Option: B
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,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

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

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

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

$f=9 cm$

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:

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

  1. $20\ cm$

  2. $60\ cm$

  3. $10\ cm$

  4. $30\ cm$

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Correct Option: D

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

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

  1. $xy$

  2. $\sqrt{xy} $

  3. $\dfrac{x+y}{2} $

  4. $\sqrt{\dfrac{x}{y} }$

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Correct Option: B

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

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

  1. $8.55 cm$

  2. $7.55 cm$

  3. $6.55 cm$

  4. $5.55 cm$

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Correct Option: A

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 

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

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

  2. $R\left(1-\dfrac {1}{2\cos \theta}\right)$

  3. $\dfrac {R}{2\cos \theta}$

  4. $\dfrac {R}{2}(1+\cos \theta)$

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Correct Option: A

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:

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

  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. $-fp$

Show answer
Correct Option: A
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: $-\dfrac{f}{p}$

A concave spherical mirror has a focal length of 12 cm. if an object is placed 6 cm in front of it, the position of the magic is 

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

  1. 4 cm behind the mirror

  2. 4 cm in front of the mirror

  3. 12 cm behind the mirror

  4. 12 cm in front of the mirror

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Correct Option: A

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

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

  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

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Correct Option: B

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

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

  1. 1 : 9

  2. 1 : 3

  3. 3 : 1

  4. 2 :1

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Correct Option: C
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 

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 :

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

  1. 1:2

  2. 3:1

  3. 1:3

  4. 2:1

Show answer
Correct Option: B
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