Question 2A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

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Accepted Answer

Given,Image distance (v) = 50 cmObject height (ho) = Image height (hi)magnification = -1 [∵ image is real & inverted]Object distance (u) = ?Power (P) = ?We know, magnification (m) = vu\dfrac{\text{v}}{\text{u}}uv​Substituting the values we get,-1 = 50u\dfrac{50}{\text{u}}u50​⇒ u = -50 cmTherefore, the needle is placed at a distance of 50 cm in front of the convex lens.According to the Lens formula,1v\dfrac{1}{v}v1​ - 1u\dfrac{1}{u}u1​ = 1f\dfrac{1}{f}f1​Substituting the values we get,150−1−50=1f⇒1f=150+150⇒1f=250⇒1f=125⇒f=25 cm⇒f=0.25 m\dfrac{1}{50} - \dfrac{1}{-50} = \dfrac{1}{f} \[0.5em] \Rightarrow \dfrac{1}{f} = \dfrac{1}{ 50} + \dfrac{1}{50} \[0.5em] \Rightarrow \dfrac{1}{f} = \dfrac{2}{50} \[0.5em] \Rightarrow \dfrac{1}{f} = \dfrac{1}{25} \[0.5em] \Rightarrow f = 25\text{ cm} \[0.5em] \Rightarrow f = 0.25\text{ m}501​−−501​=f1​⇒f1​=501​+501​⇒f1​=502​⇒f1​=251​⇒f=25 cm⇒f=0.25 mPower of lens (P) = 1f(in metres)\dfrac{1}{f \text{(in metres)}}f(in metres)1​⇒ P = 10.25=10025\dfrac{1}{\text{0.25}} = \dfrac{100}{\text{25}}0.251​=25100​⇒ P = 4DTherefore, power of lens = 4D.

Light - Reflection and Refraction — Question 2

Question 2