Question

Q.5) A concave lens has focal length of 20 cm. At what distance from the lens should a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed. What is the character of Image?






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Answer 1

According to the Question

It is given that,

• Type of Lens = Concave Lens
• Object height ,ho = 5cm
• Image distance ,v = -15cm
• Focal length ,f = -20cm

Firstly we calculate the image object distance .

By using Lens Formula.

• 1/v - 1/u = 1/f

by putting the value we get

➻ 1/-15 -1/u = 1/-20

➻ -1/15 - 1/u = -1/20

➻ -1/15 = -1/20 + 1/u

➻ -1/15 + 1/20 = 1/u

➻ -4+3/60 = 1/u

➻ -1/60 = 1/u

➻ -60 = u

➻ u = 60 cm

• Hence, the object is placed 60 cm in front of the lens .

Now, calculating magnification

• m = hi/ho = v/u

by putting the value we get

➻ hi/5 = -15/-60

➻ hi/5 = 15/60

➻ hi/5 = 1/4

➻ hi = 5/4

➻ hi = 1.25 cm

• Hence, the height of image is 1.25 .
• Nature of Image :- Image formed is Virtual and erect and height of image is smaller than the object.

Answer 2

Given information,

A concave lens has focal length of 20 cm. At what distance from the lens should a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed. What is the character of Image?

• Focal length (f) = -20 cm
• Height of object (ho) = 5 cm
• Distance of image (v) = -15 cm
• Size of image = ?
• Character of image = ?

Firstly lets calculate the distance of object by using lens formula. We know that,

• 1/v - 1/u = 1/f

Putting all values we get,

➡ 1/-15 - 1/u = 1/-20

➡ -(1/15) - 1/u = -(1/20)

➡ -1/u = - 1/20 + 1/15

➡ -1/u = - (1 × 3)/(20 × 3) + (1 × 4)/(15 × 4)

➡ -1/u = - 3/60 + 4/60

➡ -1/u = (- 3 + 4)/60

➡ -1/u = 1/60

➡ u = -60

Hence, distance of object is 60 cm.

Now, lets calculate the height of image. We know that magnification,

• m = v/u = hi/ho

Putting all values we get,

➡ m = -15/-60 = hi/5

➡ -15/-60 = hi/5

➡ 15/60 = hi/5

➡ 15/60 × 5 = hi

➡ (15 × 5)/60 = hi

➡ (1 × 5)/4 = hi

➡ 5/4 = hi

➡ 1.25 = hi

➡ hi = 1.25 cm

Hence, height of image is 1.25 cm.

Character i.e, nature of image is virtual, erect and smaller than object.