Question

A spherical mirror produces an image 48cm I. front of it, when an object is placed at a distance of 12 cm from its pole . calculate the focal length of the mirror and answer the following questions
a) identify the nature of the mirror
b)will the image be magnified or diminished?
c) state whether the image formed is real or virtual
d)will the image formed be erect or invested?
e)Draw a ray diagram showing the formation of image in the above case​

Answer 1

Explanation:

GIVEN:-

• Image Distance(v)=48cm= -48cm

(sign convention)

• Object Distance(u)=12cm= -12cm

SOLUTION:-

We know that,

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

$\frac{1}{f} = \frac{1}{ - 48} + \frac{1}{ - 12}$

$= \frac{ - 1 - 4}{48}$

$\frac{1}{f} = \frac{ - 5}{48}$

$f = \frac{ -48}{5}$

$f = - 9.6cm$

a)NATURE OF IMAGE:

Magnification(M):

${\boxed{M=\frac{-v}{u}}}$

$M=\frac{-48}{12}$

$M=-4$

Therefore,image is

• Real
• Magnified
• Inverted

b)image will be magnified,because its magnification M<1(greater than 1, irrespective of sign).

c)image formed will be real,because value of v(image distance) is -ve.

d)image formed will be inverted,because magnification M is -ve.

ADDITIONAL INFORMATION:

• Center of curvature is denoted by "C"
• Radius of curvature is denoted by "R"
• Always "$\color{red}{{\boxed{R=2f}}}$"
• Magnification,can also be written as

$\color{red}{{\boxed{M=\frac{h_{i}}{h_{o}}=\frac{-v}{u}}}}$

• The mirror used in this question is concave mirror.
• Image is formed beyond center of curvature.
• Mirror formula:

$\color{red}{{ \boxed{\frac{1}{f} = \frac{1}{v} + \frac{1}{u}}}}$

• Lens formula:

$\color{red}{{ \boxed{\frac{1}{f} = \frac{1}{v} - \frac{1}{u}}}}$

• Magnification for lens:

$\color{red}{{\boxed{M=\frac{h_{i}}{h_{o}}=\frac{v}{u}}}}$

HOPE THAT HELPS!!

(Sorry for delayed answer)

Answer 2

Given:

Image distance $(v) = 48 cm$

Object distance $(u)=-12cm$

To find:

1. Focal length $(f)$ of the mirror.
2. Type of mirror used.
3. Nature of image.

Explanation:

We have been given that the image of the object is produced in front of the mirror. Hence, image distance will be negative. $(v)=-48cm$

We have, mirror formula which states that

$\frac{1}{v}+ \frac{1}{u}= \frac{1}{f}$    $;$ $or$

$f=\frac{uv}{u+v}$

Substituting the given values in the equation, we get

$f=\frac{(-12)(-48)}{-12+(-48)}$

$f=-9.6cm$

• Since the focus of the given mirror is negative, we can conclude that the mirror is concave.

We know, the magnification of a mirror is given by

$m=\frac{-v}{u}$

Substituting the given values in the equation, we get

$m=\frac{-(-48)}{-12}$   $;$ $or$

$m=-4$

Hence,

• Magnification of the image will be -4,

• The negative sign here, indicates that the image formed by the concave mirror is real and inverted.

• Also, the image will be enlarged and magnified as compared to the object by a factor of 4.

For the ray diagram, consider the image attached below.

Final answer:

1. Focal length of the mirror is -9.6 cm.
2. The mirror used in the question is Concave mirror.
3. The image here, will be magnified by a factor of 4.
4. The image formed will be Real.
5. The image formed will be inverted at a distance 48 cm in front of the mirror.