Question

Consider an object 7.0 cm in length which is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. the size and position of image formed is

Answer 1

Answer:

of curvature (R) = 30 cm

f = R/2 = 30/2 = 15 cm

u = -20 cm,

h= 5 cm.

1/v +1/u = 1/f

1/v = 1/f – 1/u

1/v = 1/15 – 1/-20

v = 8. 57 cm

Image is virtual and erect and formed behind the mirror.

m = -v/u

m = hi/ho

hi/5= 8.57/20

= 0.428

hi = 0.428 x 5 = 2.14cm

Position of Image: Behind the mirror.

Nature of Image: Virtual and Erect.

Size of the Image: Diminished.

Step-by-step explanation:

Answer 2

Answer:

The size and position of the image formed is  $\mathrm{\frac{60}{7}\ cm}$  and 3 cm respectively.

Explanation:

We will use the mirror formula to solve this question. i.e.

$\mathrm{\frac{1}{f}= \frac{1}{v}+\frac{1}{u}}$            (1)

Where,

f=focal length of the mirror

v=image distance of the mirror

u=object distance from the mirror

From the question we have,

The height of the object(h₁)=7 cm

The object distance from the mirror(u)=-20cm

The radius of curvature of the mirror(R)=30cm

The focal length of the mirror(f)=15cm       (∵$\mathrm{f=\frac{R}{2} }$)

Now substituting the values of "f" and "u" in equation (1) we get;

$\mathrm{\frac{1}{15}= \frac{1}{v}+\frac{1}{-20}}$

$\mathrm{\frac{1}{v}=\frac{1}{15}+\frac{1}{20}}$

$\mathrm{\frac{1}{v}=\frac{4+3 }{60}}$

$\mathrm{\frac{1}{v}=\frac{7 }{60}}$

$\mathrm{v=\frac{60}{7}\ cm}$                  (2)

Now,

$\mathrm{m=\frac{-v}{u} =\frac{h_2}{h_1} }$              (3)

m=magnification of the image

h₂=image size

h₁=object size

By inserting all the required values in equation (3) we get;

$\mathrm{\frac{-\frac{60}{7}}{-20} =\frac{h_2}{7} }$

$\mathrm{ \frac{h_2}{7}=\frac{-60}{7\times -20} }$

$\mathrm{h_2=3\ cm}$         (4)

The size and position of the image formed is  $\mathrm{\frac{60}{7}\ cm}$  and 3 cm respectively.

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