Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Lenses. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Lenses Quiz 2

physics

 Rahul needs to get a perfect picture of an object four times its actual size for a physics experiment. The image is created on a screen that is $90\,cm$ away from the object. The lens used for this reason has a focal length of ______ .

Substitute the given magnification and the given condition in lens formula to arrive at the answer. Take care of sign convention. 

 $\text{Given magnification is equal to 4.}\\ \dfrac{v}{u}=4 \\  \Rightarrow v=4u\to (1) \\  \text{Also we know that}\\  v+u=90\to (2)\\  \text{From (1) and (2), we have}\\ 5u=90 \\  \Rightarrow u=18\,cm \\  \Rightarrow v=4\times 18=72\,cm \\  \text{Now, from lens formula, we have}\\ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\  \Rightarrow \dfrac{1}{f}=\dfrac{1}{72}-\dfrac{1}{-18} \\  \Rightarrow \dfrac{1}{f}=\dfrac{5}{72} \\  \Rightarrow 72=5f \\  \therefore f=12.4\,cm$ 

 During a photoshoot in Mumbai's film city, the image of an actress created by the camera's converging lens is erect.  The height of the actress in it is five times its original size.  What distance is the actress actually standing from the lens if the focal length of the camera lens is $10\,cm$?

 Compare the given heights with image/object distance using magnification formula and then, use lens equation to arrive at the answer. 

 $\text{Given magnification is equal to 5.}\\ \dfrac{v}{u}=5 \\  \Rightarrow v=5u \\  \text{Also given that}\\  f=10\,cm\\  \text{Now, from lens formula, we have}\\ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\  \Rightarrow \dfrac{1}{f}=\dfrac{1}{5u}-\dfrac{1}{u} \\  \Rightarrow \dfrac{1}{10}=\dfrac{-4}{5u} \\  \Rightarrow 5u=-40cm \\  \therefore u=-8\,cm$

 In a Business company, an employee has to give a presentation to the clients. While giving the presentation, the employee placed the slide at a distance of $15\,cm$ behind the lens of the projector. If the focal length of the converging lens is $14.8\,cm$ , then the image is obtained on a screen placed at a distance of $d\,cm$ from the lens. The value of $d$ is 

Use lens makers formula with proper sign convention to arrive at the correct answer. 

$\text{Given; u=15 cm and f=14.8 cm}\\  \text{From lens formula,}\\ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\  \Rightarrow \dfrac{1}{14.8}=\dfrac{1}{d}-\dfrac{1}{-15} \\  \Rightarrow \dfrac{1}{14.8}-\dfrac{1}{15}=\dfrac{1}{d} \\  \Rightarrow \dfrac{1}{d}=\dfrac{0.2}{14.8\times 15} \\  \therefore d=1110\,cm$  

 If $v$ is the distance where an image of an object is formed in the screen, $f$ is the focal length of the lens and $m$ is the magnification produced by the lens, then which of the following given options correctly relates $v$ , $f$ , $m$ with each other? 

Using magnification formula as well as lens formula, we can arrive at the answer. 

$\text{W.K.T magnification of lens is given by}\\  m=\dfrac{v}{u}\to (1)\\  \text{From lens formula,}\\  \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\to (2)\\  \text{Substituting (1) in (2),we have}\\ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\  \Rightarrow \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{m}{v} \\  \Rightarrow \dfrac{1}{v}-\dfrac{1}{f}=\dfrac{m}{v} \\  \Rightarrow \dfrac{f-v}{fv}=\dfrac{m}{v} \\  \Rightarrow \dfrac{1}{m}=\dfrac{f}{f-v} \\  \therefore \dfrac{1}{m}=1+\dfrac{v}{f-v}$

 In the physics laboratory, Mohan placed a wooden box and screen at a distance of $100\,cm$. Now, another student Priyanka placed a diverging lens between the box and the screen forming a sharp image of the box with magnification $3$ on screen. Now, if Mohan shifts the lens $15\,cm$ toward the screen, the sharp image is still formed intact. The focal length of the lens used for the purpose is  

Use magnification formula and the given condition to obtain initial object/image distance. From these, obtain final object/image distances to use them in the lens equation. 

At first, when Priyanka places the lens in between, the magnification is given by $ \\ \dfrac{v}{u}=3 \\  \Rightarrow v=3u \\ $ It is given that the distance between the object and the screen is $100cm.$ $ \\  $ Therefore, we can conclude that $ \\ v+u=100 \\  \Rightarrow 3u+u=100 \\  \Rightarrow 4u=100 \\  \Rightarrow u=25\,cm \\  \Rightarrow v=75\,cm \\ $ Now, when Mohan shifts the lens $15cm$ towards the screen, the image distance reduces to $v'=75-15=60\,cm$ and the object distance increases to $ u'=25+15=40\,cm$ $ \\ $ Now, focal length of diverging lens is given by $ \\ \dfrac{1}{f}=\dfrac{1}{v'}-\dfrac{1}{u'} \\  \Rightarrow \dfrac{1}{f}=\dfrac{1}{60}-\dfrac{1}{-40} \\  \Rightarrow \dfrac{1}{f}=\dfrac{1}{24} \\  \therefore f=24\,cm=0.24\,m$ 