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  4. Spherical Mirror Formula And Magnification MCQ - Practice Questions with Answers

Spherical Mirror Formula And Magnification MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Mirror formula is considered one of the most asked concept.

  • 24 Questions around this concept.

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QuestionMEDIUM

 In an experiment, a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it.  It is found that an object and its image coincide if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is :

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A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface.  A small air bubble is 6 cm below the flat surface inside it along the axis.  The position of the image of the air bubble made by the mirror is seen :

 

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 You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face and views the magnified image of the face at the closest comfortable distance of 25 cm. The radius of curvature of the mirror would then be :

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A concave mirror of focal length 'f1' is placed at a distance of 'd' from a convex lens of focal length 'f2'. A beam of light coming from infinity and falling on this convex lens - concave mirror combination returns to infinity. The distance 'd' must equal:

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Match the corresponding entries of column 1 with column 2 [where m is the magnification produced by the mirror]

Column 1

Column 2

(A)    m = -2

(a) Convex mirror
(B)    \text{m} = - \frac{1} {2}

(b) Concave mirror
(C)    m = +2

(c) Real image
(D)    \text{m} = + \frac{1} {2}

(d) Virtual image
View Solution

A thin rod of length \mathrm{f / 3}  is placed along the optical axis of a concave mirror of focal length \mathrm{f } such that its image which is real and elongated just touches the rod. Calculate the magnification.

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Concepts Covered - 4

Mirror formula

Mirror formula

Let the object distance (u), image distance (v) and focal length (f). 

The following sign convention is used for measuring various distances in the ray diagrams of spherical mirrors:

  • All distances are measured from the pole of the mirror.
  • Distances measured in the direction of the incident ray are positive and the distances measured in the direction opposite to that of the incident rays are negative.
  • Distances measured above the principal axis are positive and that measured below the principal axis are negative.

 

\begin{array}{l}{\text { In } \Delta A B C \text { and } \Delta A^{\prime} B^{\prime} C^{\prime}} \\ \\ {\Delta A B C \sim \Delta A^{\prime} B^{\prime} C[\text { AA similarity }]} \\ \\ {\frac{A B}{A^{\prime} B^{\prime}}=\frac{A C}{A^{\prime} C^{\prime}} \ldots(1)}\end{array}

 

Similarly, In \triangle{FPE} and \triangle{A'B'F'} 

\begin{array}{l}{\frac{E P}{A^{\prime} B^{\prime}}=\frac{P F}{A^{\prime} F}} \\ \\ {\frac{A B}{A^{\prime} B^{\prime}}=\frac{P F}{A^{\prime} F}[\mathrm{AB}=\mathrm{EP}] \cdots(\mathrm{II})}\end{array}

From (i) &(ii) 

 \begin{array}{l}{\frac{A C}{A^{\prime} C}=\frac{P F}{A^{\prime} F}} \\ \\ {=>\frac{A^{\prime} C}{A C}=\frac{A^{\prime} F}{P F}} \\ \\ {=>\frac{\left(C P-A^{\prime} P\right)}{(A P-C P)}=\frac{\left(A^{\prime} P-P F\right)}{PF} }\end{array}

 

Now, PF = -f  ;  CP=2PF=-2f ; AP= -uAP=-u \ \text{and} \ A'P=-v

Put these value in above relation: 

\begin{array}{l}{\Longrightarrow \frac{[(-2 f)-(-v)]}{(-u)-(-2 f)}=\frac{[(-v)-(-f)]}{(-f)}} \\ \\ {\Longrightarrow u v=f v+u f} \\ \\ {\Longrightarrow \frac{1}{f}=\frac{1}{u}+\frac{1}{v}}\end{array}

Proved.

Magnification in Spherical mirrors

Magnification in Spherical mirrors: 

lateral magnification: 

The lateral magnification is defined as the ratio: 
    m_{v}=\frac{\text { height of image }}{\text { height of object }}=\frac{h_{i}}{h_{0}}

To compute the vertical magnification, consider the extended object OA shown in Figure. The base of the object, O will map
on to a point I on the principal axis which can be determined from the equation    \frac{1}{u}+\frac{1}{v}=\frac{1}{f}  .

The image of the top of the object A, will map on to a point A' that will lie on the perpendicular through I. The exact location can be determined by drawing a ray from A passing through the pole and intercepting the line through I at A'.

Consider the triangles APO and A'PI  in the figure. As the two
triangles are similar, we get, 
\tan \alpha=\frac{A O}{P O}=\frac{A^{\prime} I}{P I} \quad \text { or } \quad \frac{A^{\prime} I}{A O}=\frac{P I}{P O}

Applying the sign convention, we get, u=-PO
v=-P I \Rightarrow h_{0}=+A O \quad \Rightarrow \quad h_{i}=-A^{\prime} Iv=-P I \Rightarrow h_{0}=+A O \quad \Rightarrow \quad h_{i}=-A^{\prime} I


Therefore, \quad-\frac{h_{i}}{h_{0}}=\frac{v}{u} \quad$ or $\quad m_{v}=\frac{h_{i}}{h_{0}}=-\frac{v}{u}

magnification formula can be modified as: 
$$ m=\frac{-v}{u}=\frac{f }{f -u} $$=\frac{f-v}{f}

Longitudinal magnification:  When object lies along the principal axis then its axial magnification 'm' is given by

                                            m=\frac{I}{O}=\frac{-\left(v_{2}-v_{1}\right)}{\left(u_{2}-u_{1}\right)}

If the object is small, 

               m=-\frac{d v}{d u}=\left(\frac{v}{u}\right)^{2}=\left(\frac{f}{f-u}\right)^{2}=\left(\frac{f-v}{f}\right)^{2}

 

 

 

 

 

 

 

 

Relation between velocity of object and mirror in Spherical mirror

Relation between velocity of object and mirror in Spherical mirror

Case I: when the object moves along principal axis

When we differentiate equation   \frac{1}{v}+\frac{1}{u}=\frac{1}{v}   with respect to time.

\begin{array}{ll}{\Rightarrow \quad} & {-\frac{1}{v^{2}} \frac{d v}{d t}-\frac{1}{u^{2}} \frac{d u}{d t}=0} \\ \\ {\Rightarrow \quad} & {-\frac{1}{v^{2}} V_{i m}-\frac{1}{u^{2}} V_{O M}=0}\end{array}

              \frac{d v}{d t}=v_{i m}    = velocity of image w.r.t. miror

\Rightarrow \quad V_{i m}=-\frac{v^{2}}{u^{2}} V_{O M}

  \frac{d u}{d t}=v_{o M}   =  velocity of object w.r.t. mirror

V_{i m}=-m^{2} V_{O M}

\implies V_{i }-V_{m}=-m^{2} (V_{O }-V_{m})

Therefore, \implies V_{i }=-m^{2} V_{O }  when mirror is at rest along the principal axis.

Case II: when the object moves perpendicular to  principal axis

When an Object is moving perpendicular to the principal axis. This time v (image distance) and u (object distance) are constant. 

Therefore we have the following relation :

\frac{h_{i}}{h_{o}}=\frac{v}{u} \ \text{or} \ h_{2}=\left(\frac{-v}{u}\right)h_{1}

Also, x-coordinates of both image and object remain constant. On differentiating the above relation w.r.t time we get, 

\frac{d h_{i}}{d t}=-\frac{v}{u} \frac{d h_{o}}{d t}

Here, \frac{d h_{o}} {dt}  denotes velocity of object perpendicular to the principal axis and \frac{d h_{i}} {dt} denotes velocity of the image perpendicular to the principal axis.

Hence we can conclude that, 

\frac{d h_{i}}{d t}=-\frac{v}{u} \frac{d h_{o}}{d t}

\implies (V_{im})_y= \frac{-v}{u}(V_{om})_y \\

\implies (V_{im})= m(V_{om})_y \\

Newton's Formula

Newton's Formula:

As we know that the mirror formula is given as 

\frac{1}{v}+\frac{1}{u}=\frac{1}{f}

Let's assume, x = distance of the object from focus 

                       y =  distance of the image from focus 

Newton's formula is useful  for calculating the image position for a curved mirror.

The diagram shows the position of an object and its image formed by a concave mirror.

Let the distances of the object and image from the principal focus of the mirror be x and y respectively.

Then: Object distance (u) = f+x and Image distance (v) = f+y

Using the mirror formula \frac{1}{v}+\frac{1}{u}=\frac{1}{f}  we have: \frac{1}{f+x}+\frac{1}{f+y}=\frac{1}{f}   

and simplifying this we get:    \boxed{f^2= xy }

 

Study it with Videos

Mirror formula
Magnification in Spherical mirrors
Relation between velocity of object and mirror in Spherical mirror

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