Question

a security mirror used in big showroom has radius of curvature 5 metre if a customer is standing at a distance of 20 m from the cash counter find the position nature and size of image formed in the security mirror ​

Answer 1

Question :-

A security mirror used in big showroom has radius of curvature 5 metre if a customer is standing at a distance of 20 m from the cash counter find the position nature and size of image formed in the security mirror .

Answer :-

→ Image distance (v) = 2.11 m

→Nature of image is virtual,erect and diminished

To Find :-

→ Position ,nature and Size of image

Explanation :-

We know that convex mirror is used in security mirrors -

According to the question -

→ Customer is standing at a distance of 20 m ( object distance ) and radius of curvature of mirror is 5 m

$\pink{\bf{Given}\begin{cases}\sf{ \green{object \: distance \: (u) = 20 \: m \:} }\\\sf{ \purple{radius \: of \: curvature( R)= 5m}}\\ \sf{ }\end{cases}</p><p> \: }$

Let, image distance is "v"

We know that -

For convex mirror object distance should be negative for observation ,and focal length also positive -

Hence ,apply -

$\mapsto \sf \red{ \frac{2}{R} = \frac{1}{v} + } \frac{1}{u} \\ \: \\ \mapsto \sf \green{ \frac{2}{5} = } \frac{1}{v} \pink{ - \frac{1}{20} } \\ \\ \mapsto \sf \blue{ \frac{1}{v} =} \frac{2}{5} + \green{ \frac{1}{20}} \\ \\ \mapsto \sf \frac{1}{v} \red{ = \frac{8 + 1}{20} }\\ \\ \mapsto \sf \frac{1}{v}\orange{ = \frac{9}{20} } \\ \\ \mapsto \sf \: v = \pink{\frac{20}{9}} \\ \\ \mapsto \boxed{ \sf \: v = 2.11 \: m}$

°•° Image distance (v) is positive hence image formed on the same side of mirror as object .

Nature of image is erect , virtual ,small and diminished .

→ °•° Magnification will 1/9th of size of customer hence image should be smaller then object (customer) .

Answer 2

$\huge\tt{Answer:}$

Given:

➹A security mirror used in big showroom has radius of curvature 5 metre.

➹ If a customer is standing at a distance of 20 m from the cash counter.

Find:

➹ Find the position nature and size of image formed in the security mirror .

Note for the questionor:

➹ Let, "V" image distance.

➹ Convex mirror object distance should be negative.

➹ Focal length ait should be positive.

Know terms:

$\textsf {Here} \begin{cases} (u) = \textsf{Initial velocity} \\ (v) = \textsf{Final Velocity} \\ (f) = \textsf{Focal length} \\ (R) = \textsf{Radius} \end{cases}$

$\sf= \frac{2}{R} = \frac{1}{v} + \frac{1}{u} \\ \sf = \frac{2}{5} = \frac{1}{v} - \frac{1}{20} \\ \sf = \frac{1}{v} = \frac{2}{5} + \frac{1}{20} \\ \sf = \frac{1}{v} = \frac{8 + 1}{20} \\ \sf = \frac{1}{v} = \frac{9}{20}$

$\boxed{\boxed{\sf v=\frac{20}{9}}}}$

$\sf v = 2.8\: m$

Therefore, the distance of images is negative and other side of mirror object and Nature of image is virtual and opposite right side will be erect. hence they are virtual and erect.