Question

1. A spherical mirror produces an image of magnification -1 on a screen placed at a distance of 50 cm from the mirror. (a) Write the type of mirror. (b) Find the distance of the image from the object. (c) What is the focal length of the mirror?

Answer 1

$\sf\color{red}{1.}$It is a concave mirror.

$\sf\color{red}{2.}$ Distance of the image from the object is zero

$\sf\color{red}{3.}$ Focal length= - 25cm.

$\underline{\boxed{\sf\purple{Given:}}}$

Magnification (m) = -1.

Image distance(v)= 50 cm= -50cm (by sign convention).

$\underline{\boxed{\sf\purple{To\: Determine:}}}$

$\sf\color{red}{1.}$ Type of mirror

$\sf\color{red}{2.}$ Distance of the image from the object

$\sf\color{red}{3.}$ Focal length.

1. Negative sign of magnification denotes that the image formed is inverted. Also, since the object is obtained on a screen, So, it is real. This means that the object is real and inverted. Hence, the given mirror is a concave mirror.

2. By definition of magnification we get:

$\begin{gathered}m = - \frac{v}{u} \\ \end{gathered}$

$\underline{\boxed{\sf\purple{Putting \:the \:values,\: we \:get:}}}$

$\begin{gathered} - 1 = - \frac{ - 50}{u} \\ u = - 50.\end{gathered}$

$\underline{\boxed{\sf\purple{Object \:distance \:(u)= \:-50 cm.}}}$

$\bold{Therefore,}$

distance of the image from the object= 0(since the object and the image are at the same distance from the pole).

3. By mirror formula we get:

$\begin{gathered} \frac{1}{f } = \frac{1}{v} + \frac{1}{u} \\ \frac{1}{f} = \frac{1}{ - 50} + \frac{1}{ - 50} = \frac{ - 1}{25 } \\ f = - 25.\end{gathered}$

$\bold\color{red}{Focal\: length\: (f)\:=-25cm.}$

─━─━─━─━─━─━─━─━─━─━─━─━─

Thankyou :)

Answer 2

To solve this question easily, one needs to remember the direct result of finding image position based on object position.

According to that,

• If object is at infinity, then image is formed at F.
• If object is beyond C, image is formed between F & C.
• If object is at C, then image is also at C.
• If object is between F & C, then image is beyond C.
• If object is at F, image is formed at infinity.
• If object is between P & F, then image is formed behind the mirror.
• The first 5 cases refer to the formation of a real image. The 6th case shows the formation of a virtual image.
• According to the question, magnification is equal to -1. Here negative sign implies, image is real and inverted while value of 1 implies object and image are of the same size. Real image is formed only by a concave mirror. Hence the type of mirror used is Concave.
• By the above cases, we get to know that the image and object have the same distance only if they both are placed at the center of curvature (C).

Hence the distance between object and image is zero units. (v-u = 0cm)

Now since it is given that object is placed at a distance of 50 cm, we can say that the value of C is 50 cm. We also know that, C = 2f. Hence the focal length (f) of the mirror is:

$\sf f = \frac{c}{2} = \frac{52}{2} = 25cm$

HENCE -:

$\sf Focal\;length\;(f)=−25\;cm.$

• Since the mirror is a concave mirror, the focal length is taken as -25 cm due to sign convention.

For the ray diagram, please refer to the attachment.

(Also in case you need to know about the direct results, please go through the second file being attached.)

Thanks!! (;