Question

Spherical mirrors. Object O stands on the central axis of a spherical mirror. For this situation object distance is ps = +16 cm, the type of mirror is concave, and then the distance between the focal point and the mirror is 11 cm (without proper sign).

Find :

(a) the radius of curvature r (including sign),

(b) the image distance i, and

(c) the lateral magnification m. Also, determine whether the image is

(d) real or virtual ,

(e) inverted from object O or non inverted , and

(f) on the same side of the mirror as O or on the opposite side.

Answer 1

Answer:

a

The radius of curvature is $R= +22cm$

b

the image distance $I= +35cm$

c

the lateral magnification is $m = -3.182$

d

The image is real given that the object distance is greater than the focal length and this is a type of setup that gives a real image in concave mirror

e

This would be inverted because from above we see that the image is real and all real image are inverted i concave mirror

f

This image would be on the same side of the object this is because both the image distance and the object distance are positive

Explanation:

From the question we are told that

                 The object distance is $P_s = +16cm$  

                  The focal length is  $F_c = 11cm$

The focal length is mathematically given as

                     $F_c = \frac{R}{2}$

   Where R is the radius of curvature

Now making R the subject of the formula above

             $R = 2F_c$

                 $= 2 *11 = +22cm$ (the positive sign is because it is in the real world size)

The mathematical representation of the mirror formula is

                  $\frac{1}{I} + \frac{1}{u} + \frac{1}{F_c}$

                $\frac{1}{I} + \frac{1}{16} = \frac{1}{11}$

               $\frac{1}{I} = \frac{1}{11} - \frac{1}{16}$

               $\frac{1}{I} = \frac{5}{176}$

                 $= \frac{176}{5}$

                 $I = +35.2 cm$

Mathematically  lateral magnification is represented as

             $m = \frac{I}{u}$

Where I is the image distance and u is the object distance

      $m= -\frac{35}{16} = -3.182$      

                           

Answer 2

a)  The radius of curvature R is +22 cm

b)  the image distance I is +25 cm

c)  the lateral magnification m is -3.182

d) The image is real given that the object distance is greater than the focal length and this is a type of setup that gives a real image in concave mirror

e) The image is inverted because from object O.

f )The image would be on the same side of the object this is because both the image distance and the object distance are positive

Let's solve the question:

In the question it is given that, the type of mirror is concave mirror;

Also given : Object distance = +16 cm  

The focal length = +11 cm  

The focal length is mathematically given as:

$F=\frac{R}{2}$

where  F= focal length and R =radius of curvature

So, calculation of R is as follows:

$R=2*11= +22 cm$   (the positive sign due to sign conventions)

The mathematical representation of the mirror formula is

$\frac{1}{F} = \frac{1}{I} + \frac{1}{u} \\\\\frac{1}{I} =\frac{1}{11} -\frac{1}{16} \\\\I=\frac{176}{5} \\\\I=+32.5 cm$

Mathematically, lateral magnification is represented as

$m=\frac{I}{u}$

where I =image distance and u = object distance

$m=-\frac{35}{16} =3.812$

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