Question

A convex mirror used for rear view on an automobile has a radius of curvature f 3.00m If a bus is located at 5.00m from this mirror .Find the position,nature and size of mirror.?
Please help me friends !!!!please

Answer 1

Answer:

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Given:-

Radius of curvature (R)= +3.00m

Object distance (u)= -5.00m

To find:-

Image distance (v)=?

Height of Distance(h')=?

Formulas to be used:-

Mirror formula = 1/v+1/u=1/f

Magnification= h'/h

Solution:-

Since ,we know that

Focal length(F)= R/2

=>+3.00/2=+1.5m

=>1.5m

[it is positive as principle focus of a convex mirror is behind the mirror ]

Now,

1/v+1/u=1/f

the formula becomes;

1/v= 1/f -1/u

=>+1/1.50-1/(-5.00)

=>1/1.50 +1/5.00

=>5.00+1.50/7.50

v=+1.15m

The image is 1.15m at back of mirror.

Magnification(m)=h'/h

=>-v/u

=>1.15/-5.00

Magnification=>+0.23

The image will be virtual ,erect and smaller in size by a factor of 0.23(+)

Be brainly⭐

Answer 2

Explanation:

heya

Given:-

Radius of curvature (R)= +3.00m

Object distance (u)= -5.00m

To find:-

Image distance (v)=?

Height of Distance(h')=?

Formulas to be used:-

Mirror formula = 1/v+1/u=1/f

Magnification= h'/h

Solution:-

Since ,we know that

Focal length(F)= R/2

=>+3.00/2=+1.5m

=>1.5m

[it is positive as principle focus of a convex mirror is behind the mirror ]

Now,

1/v+1/u=1/f

the formula becomes;

1/v= 1/f -1/u

=>+1/1.50-1/(-5.00)

=>1/1.50 +1/5.00

=>5.00+1.50/7.50

v=+1.15m

The image is 1.15m at back of mirror.

Magnification(m)=h'/h

=>-v/u

=>1.15/-5.00

Magnification=>+0.23

The image will be virtual ,erect and smaller in size by a factor of 0.23(+)

i hope it's help uh