Question

An object 2cm tall is placed in front of a concave mirror of radius of curvature 20cm at a distance of 5cm. Find the position, nature and size of the image.

Answer 1

According to the Question

It is given that,

• Height of object ,ho = 2cm
• Radius of Curvature ,R = 20cm
• Object distance ,u = -5cm
• Type of Mirror = Concave Mirror

we have to calculate the position , nature and size of the mirror .

Firstly we calculate the focal length of the mirror.

• f = R/2

➻ f = 20/2 cm

➻ f = 10cm

Now, calculating the image distance .

By using Mirror Formula .

• 1/v + 1/u = 1/f

Substitute the value we get

➻ 1/v + 1/-5 = 1/10

➻ 1/v - 1/5 = 1/10

➻ 1/v = 1/10 + 1/5

➻ 1/v = 1+2/10

➻ 1/v = 3/10

➻ v = 10/3 = 3.33 cm

• Hence, the image distance is 10/3 cm from the mirror.

Now, calculating the magnification of the mirror

• m = hi/ho = -v/u

➻ hi/2 = -10/3 × 1/-5

➻ hi/2 = 2/3

➻ hi = 4/3 cm

• Hence, the height of the image is 4/3 cm
• Nature of Image is Virtual , erect and smaller than the object.

Answer 2

Given :-

An object 2cm tall is placed in front of a concave mirror of radius of curvature 20cm at a distance of 5cm. Find the position, nature and size of the image.

• Height of object (ho) = 2cm
• Radius of curvature (R) = 20cm
• Distance of object = -5cm

Here, we have to find the position, nature and size of the image.

Firstly finding focal length. As we know that;

• f = R/2

Putting all values we get,

➬ f = 20/2

➬ f = 10cm

• Hence, focal length is 10cm.

Now, let's calculate the distance of image. Using mirror formula. As we know that;

• 1/v + 1/u = 1/f

Putting all values we get,

➬ 1/v + 1/-5 = 1/10

➬ 1/v = 1/10 - 1/-5

➬ 1/v = 1/10 + 1/5

➬ 1/v = 1/10 + (1 × 2)/(5 × 2)

➬ 1/v = 1/10 + 2/10

➬ 1/v = (1 + 2)/10

➬ 1/v = 3/10

➬ v = 10/3

➬ v = 3.33cm

• Hence, distance of image from mirror is 3.33cm.

Now, let's calculate magnification of mirror. As we know that;

• m = hi/ho = -v/u

Putting all values we get,

➬ hi/2 = -10/3 × 1/-5

➬ hi/2 = 10/3 × 1/5

➬ hi/2 = (10 × 1)/(3 × 5)

➬ hi/2 = 10/15

➬ hi/2 = 2/3

➬ hi = 2/3 × 3

➬ hi = (2 × 2)/3

➬ hi = 4/3

➬ hi = 1.33cm

• Hence, height of image is 1.33cm.

• Nature of image is virtual, erect and smaller than object.

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