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CBSE Class 10 Science Question 10 of 31

Light - Reflection and Refraction — Question 10

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Question 10

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed.

Answer

Given,

Object height (ho) = 5 cm

Object distance (u) = -25 cm

focal length (f) = 10 cm

Ray diagram is shown below:

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed. NCERT Class 10 Science CBSE Solutions.

Image distance (v) = ?

Image height (hi) = ?

According to the Lens formula,

1v\dfrac{1}{v} - 1u\dfrac{1}{u} = 1f\dfrac{1}{f}

Substituting the values we get,

1v−1−25=1101v+125=110⇒1v=110−125⇒1v=5−250⇒1v=350⇒v=16.66 cm\dfrac{1}{v} - \dfrac{1}{-25} = \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{25} = \dfrac{1}{10} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 10} - \dfrac{1}{25} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{5-2}{50} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{3}{50} \\[0.5em] \Rightarrow v = 16.66 \text{ cm}

Therefore, the distance of the image is 16.66 cm on the opposite side of the lens.

Magnification (m) = vu\dfrac{\text{v}}{\text{u}} = height of imageheight of object\dfrac{\text{height of image}}{\text{height of object}}

Substituting the values we get,

16.66−25\dfrac{16.66}{-25} = height of image5\dfrac{\text{height of image}}{5}

height of image = 16.66×5−25\dfrac{16.66 \times 5}{-25} = -3.3

Negative sign of the height of image means that an inverted image is formed.

The image is reduced, real and inverted.