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This is second of our many quizzes on Geometric optics. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Geometric optics Quiz 2

physics

The axial magnification of a hair pin placed along the axis of a mirror is equal to-$(\text{Take the distance of it from the mirror to be u.} \\ \text{And the corresponding image distance to be v})$ 

 Differentiate the mirror formula and obtain an expression for axial magnification. 

 $\text{In this case, different points of the hair pin are at different distances from the mirror.}\\ \text{Let us take}\,\Delta u\,\text{to be the length of the hair pin}\\ \text{and}\,\Delta v\,\text{to be the length of image of the hair pin and f to be a constant.}\\ \text{From mirror formula, we have}\\ \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\ \text{Differentiating the above expression, we have}\\-\dfrac{\Delta v}{{{v}^{2}}}-\dfrac{\Delta u}{{{u}^{2}}}=0 \\  \Rightarrow \dfrac{\Delta v}{\Delta u}=\dfrac{{{v}^{2}}}{{{u}^{2}}} \\ \therefore \dfrac{\Delta v}{\Delta u} ={{m}^{2}} \\ \text{Here, we can conclude that axial magnification is equal to the square of linear magnification.}$ 

A shining pearl is placed at the bottom of a tall cylindrical jug. How much water of refractive index $\dfrac{4}{3}$ must be added to the jug so that its image appears nearer by $10\text{ cm}$? 

 Using the formula for normal shift which connects both refractive index and thickness of water, the required depth can be determined.  

  $\text{When the pearl is viewed through water, we have the formula for normal shift as} \\x=\dfrac{(\mu -1)t}{\mu } \\  \Rightarrow t=\dfrac{\mu x}{\mu -1} \\  \Rightarrow t=\dfrac{\dfrac{4}{3}\times 10}{\dfrac{4}{3}-1}\\  \therefore t =40\text{ cm} \\ \text{Clearly, water must be poured to a depth of}\,40\text{ cm}.$

Fill up using the diagram as reference.$\\$ Deviation produced by ____ is given by $180-\_\_\_\_=2\theta$ $\\$ Deviation produced by ____ is given by $i-r$ or ____ $\\$ The total deviation produced by successive reflections from two mirrors is given by $360-\_\_\_\_$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6c33f3f3-aff7-4938-8391-79aa420359dd9194006392583088777.png"/> 

 Recall the mathematical expressions for deviations caused by refraction, reflection and successive reflections from two mirrors. 

$\text{From the given diagram as well as known facts, we have:} \\ \text{Deviation produced by reflection is given by}\, 180-2i=2\theta.\\ \text{Deviation produced by refraction is given by}\, i-r\,or\, r-i.\\ \text{The total deviation produced by successive reflections from two mirrors is given by}\,360-2\theta.$ 

 Saniya is enjoying a ride on a hot air balloon in Turkey. When she reached a height of about $4\text{ km}$, she takes out her film-camera of focal length $10\text{ cm}$ to click a picture of a rectangular farm plot of length $2\text{ km}$ and breadth $1\text{ km}$. Evaluate the minimum length and the minimum breadth of the film required for the whole plot to come in the picture.  

 An expression for magnification of a camera lens is derived by equating it to the length of object as well as length of image. Note that image distance in this case is nothing but the focal length of the lens. 

$\text{In this case, the object distance is very large.}\\ \text{Hence, the image is practically formed at the focus of the camera lens.}\\ \text{Magnification can be expressed as}\\\dfrac{v}{u}=\dfrac{{{l}_{image}}}{{{l}_{object}}} \\   \Rightarrow \dfrac{10\text{ cm}}{4\text{ km}}=\dfrac{{{l}_{image}}}{2\text{ km}} \\  \Rightarrow {{l}_{image}}=5\text{ cm} \\ \text{Clearly, the minimum length and the minimum breadth of the film required}\\ \text{for the whole plot to come in the picture are}\,5\text{ cm}\,\text{and}\,2.5\text{ cm}\,\text{respectively.}$ 

For a terrestrial telescope of length $L$, if the small focal length of the eyepiece is given by ${{f}_{E}}$ then the large focal length of objective ${{f}_{O}}$ for normal adjustment is given by $(\text{Take}\,\alpha\,\text{to be the angle subtended by the distant object}\\ \text{and}\,\beta\,\text{to be the angle subtended by its image on the eye})$ 

 Recall the expressions for magnifying power of a terrestrial telescope for normal adjustment as well as its length and connect them both. 

 $\text{We know that length of a terrestrial telescope is given by}\\ L={{f}_{O}}-{{f}_{E}} \\ \text{Also do we know that magnifying power of this telescope for normal adjustment is given by}\\ \dfrac{{{f}_{O}}}{{{f}_{E}}}=\dfrac{\beta }{\alpha } \\\Rightarrow \dfrac{{{f}_{O}}}{{{f}_{O}}-L}=\dfrac{\beta }{\alpha } \\ \Rightarrow \alpha {{f}_{O}}=\beta ({{f}_{O}}-L) \\ \Rightarrow {{f}_{O}}(\alpha -\beta )=-\beta L \\ \therefore {{f}_{O}}=\dfrac{-\beta L}{\alpha -\beta } $