Question

the radius of curvature of a convex mirror used on a moving automobile is 2.0 m. A truck is coming behind it at a constant distance of 3.5 m. Calculate (1) position of image (2) size of image relative to size of truck (3) find nature of the image formed​

Answer 1

Given:

• Radius Of Curvature of a Convex Mirror(R) = 2.0 m = 200 cm.
• Object Distance(u) = 3.5 m = 350 cm.

Find:

1. Position of Image(v).
2. Relative Size of image, i.e. Magnification(m).
3. Nature of the image formed.

Solution:

We Know That:

R = 2f. [Where R is the Radius of Curvature of the Mirror and f is the focal length of the mirror].

f = R/2.

f = 200/2.

f = 100 cm.

Applying Sign Convention:

f = +100 cm. [Focal length of Convex mirror is always positive].

u = -350 cm.

According to Mirror Formula:

1/f = 1/v + 1/u.

1/100 = 1/v + 1/-350.

1/100 = 1/v - 1/350.

1/v = 1/100 + 1/350.

1/v = 7/100 + 2/700.

1/v = 9/700.

v = 700/9.

v = +77.77 cm.

v = +0.77 m.

Magnification formula:

m = -v/u.

m = -700/9 * -1/350.

m = +2/9.

m = +0.22.

Since the value of the position of the image(v) is positive, the image will be formed on the non-reflecting side of the mirror, which is, behind the mirror.

The nature of such images is virtual and erect. The image is said to be diminished.

Therefore:

1. Position of the image(v) = +0.77 m.
2. Magnification(m) = +0.22 m.
3. Nature of the image formed = Virtual, Erect and Diminished.

Answer 2

♣ Given :

Radius of Curvature (R) = 2.0 m

Object Distance (u) = 3.5 m

♣ To Find

Position of Image

Size of image (truck image)

Nature of image formed

♣ Solution :

As we know that :

$\large {\mathrm{\boxed{R \: = \: 2f}}}$

Where, R is radius of curvature and f is focal length.

Put values

$\rightarrow{f \: = \: \dfrac{R}{2}}$

$\rightarrow {f \: = \: \dfrac{2}{2}}$

$\rightarrow {focal \: length \: (f) \: = \: 1 \: m}$

•°• Focal length is 1 m

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Now use formula :

$\large{\boxed{\rm{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: + \: \dfrac{1}{u}}}}$

$\rightarrow {\dfrac{1}{f} \: = \: \dfrac{1}{v} \: + \: \dfrac{1}{-3.5}}$

$\rightarrow {\dfrac{1}{f} \: = \: \dfrac{1}{1} \: - \: \dfrac{1}{3.5}}$

$\rightarrow {\dfrac{1}{v} \: = \: \dfrac{1}{1} \: + \: \dfrac{1}{3.5}}$

$\rightarrow {\dfrac{1}{v} \: = \: \dfrac{ 3.5 \: + \: 1}{3.5}}$

$\rightarrow {\dfrac{1}{v} \: = \: \dfrac{4.5}{3.5}}$

$\rightarrow {\dfrac{1}{v} \: = \: \dfrac{9}{7}}$

$\rightarrow {v \: = \: \dfrac{7}{9}}$

$\implies {\boxed{\sf{Image \: Distance \: (v) \: = \: 0.77 \: m}}}$

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Formula for Magnification is :

$\large{\boxed{\rm{m \: = \: - \: \dfrac{v}{u}}}}$

$\rightarrow {m \: = \: - \: \dfrac{0.77}{-3.5}}$

$\rightarrow {m \: = \: \dfrac{0.77}{3.5}}$

$\implies {\boxed{\sf{m \: = \: 0.22 \: D}}}$

• Magnification is +0.22 D

• Nature is Virtual and Erect

• Size is Diminished