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

Vision Quiz 2

physics

The following diagram shows an eye and the eye lens. The height of the object is $4\text{ cm}$ and the height of the image is $4\text{ mm}$ . Find the distance $x$ if the focal length of the eye lens is $1.5\text{ mm}$. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/584cc169-87cf-460e-b577-0e0481cf32ea2022310370557642598.png"/>

 The negative ratio of the image distance to the object distance from the lens is the magnification of the lens.

The magnification due to a convex lens is given as; $m=\dfrac{-v}{u}$ The magnification can also be given as; $m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$ Substituting the values; $\dfrac{{{h}_{i}}}{{{h}_{o}}}=-\dfrac{v}{x} \\ \Rightarrow \dfrac{4}{40}=-\dfrac{1.5}{x} \\ \Rightarrow x=\dfrac{-1.5\times 40}{4} \\ \therefore x=-15\text{ mm} $

  In case of a hypothetical eye that has a diameter of $30\text{ mm}$ , if a pencil is kept at a distance of $120\text{ mm}$ and the image is formed at $12\text{ mm}$ from the lens, what can you conclude from this observation about the person having such an eye?

  In order to have the  image of an object the image must be formed on the retina of the eye. If it is formed in front of the retina or behind the retina, the image will be blurred.

From the lens formula; $\dfrac{1}{u}-\dfrac{1}{v}=\dfrac{1}{f} \\ \Rightarrow \dfrac{1}{120}-\dfrac{1}{12}=\dfrac{1}{f} \\ \Rightarrow \dfrac{1}{f}=\dfrac{1-10}{120} \\ \therefore f=\dfrac{120}{-9}=\dfrac{-40}{3}=-13.33$ Given that $D=30\text{ mm}$. Therefore, $f < D$. But ideally for a normal eye $f=D$. Therefore, the Image of this object and all other objects will be formed in front of the retina and not on the retina. This happens in the case of myopia. So, this person has Myopia.

  A defective eye sees an object placed at $160\text{ mm}$ from the eye lens. The image of this object is blurred. What must be the focal length of the lens that is used to have a clear image if the lens is $20\,\text{mm}$ away from the eye lens? Assume the lens to be convex.

 The artificial convex lens and the eye lens together behave as a lens displacement apparatus. For clear image the distance between the lenses is directly proportional to the distance of the object from the screen.

In order for the image of the object to be clear; $d=\sqrt{{{D}^{2}}-4Df}$ But the diameter of a human eye is approximately $25\text{ mm}$ so distance of the screen will be; $D=160+{{d}_{eye}}=160+25 \\ \Rightarrow D=185\text{ mm} \\ $ Substituting the values; $d=\sqrt{{{D}^{2}}-4Df} \\ \Rightarrow d=\sqrt{D\left( D-4f \right)} \\ \Rightarrow 20=\sqrt{185\left( 185-4f \right)} \\ \Rightarrow 400=185\left( 185-4f \right) \\ \Rightarrow 4f=185-\dfrac{80}{37}=\dfrac{6845-80}{37} \\ \therefore f=\dfrac{6765}{37\times 4}=45.71\,\text{mm} $

  A man is suffering from the problem of farsightedness. The minimum distance he can see with his bare eyes is $150\text{ cm}$ . He then uses a lens of power $5\text{ D}$ to see an object placed at $50\text{ cm}$ from the eye. Determine if this object is visible to the man using that lens.

 The man will be able to see the object clearly only when the image of the object is formed on the retina. The lens formula will come in handy to check the values given.

The focal length of the lens is given as; $f=\dfrac{-1}{P}=\dfrac{-1}{5}=-0.2\text{ m} \\ \Rightarrow f=-20\text{ cm} $ From the lens formula; $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \\ \Rightarrow \text{LHS }=\dfrac{1}{-20} \\ \Rightarrow \text{RHS }=\dfrac{1}{-150}-\dfrac{1}{-50} \\ \Rightarrow \text{RHS }=\dfrac{1}{50}-\dfrac{1}{150} \\ \Rightarrow \text{RHS }=\dfrac{3-1}{150}=\dfrac{2}{150}=\dfrac{1}{75} \\ \therefore \text{RHS}\ne \text{LHS} \\ $ The equation does not satisfy this focal length. This means that the man will not be able to see the object with the help of the given lens.

 For an aging man the focal length of his eye lens changes with his age as $f\left( y \right)=4y+2$ . Here $f$ is the focal length and $y$ is the age of the man in years. Assuming the man to be 24 years old, calculate the power of the lens used to see an object $100\text{ mm}$ away from the eye lens. Assume the distance between the convex lens used and the eye to be $25\text{ mm}$ and the diameter of the eye to be $25\text{ mm}$.

 The convex lens and the eye lens will behave like a two-lens system where the lenses are separated by a finite distance between them. The effective focal length of this system can be calculated as the object and Image distances are mentioned.

Consider the effective focal length of this two-lens system separated by some finite distance to be ${{f}_{eff}}$ From the lens formula; $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{{{f}_{eff}}} \\ \Rightarrow \dfrac{1}{-25}-\dfrac{1}{-100}=\dfrac{1}{{{f}_{eff}}} \\ \Rightarrow \dfrac{1}{{{f}_{eff}}}=\dfrac{1-4}{100}=\dfrac{-3}{100} \\ $ But the effective focal length is given by; $\dfrac{1}{{{f}_{eff}}}=\dfrac{1}{{{f}_{1}}}+\dfrac{1}{{{f}_{2}}}-\dfrac{d}{{{f}_{1}}{{f}_{2}}}$ Given that ${{f}_{2}}=4y+2=4\times 24+2 \\ \Rightarrow {{f}_{2}}=98\text{ mm} \\ $ Substituting the values; $\dfrac{-3}{100}=\dfrac{1}{{{f}_{1}}}+\dfrac{1}{98}-\dfrac{25}{98{{f}_{1}}} \\ \Rightarrow \dfrac{-3}{100}-\dfrac{1}{98}=\dfrac{1}{{{f}_{1}}}\left( 1-\dfrac{25}{98} \right) \\ \Rightarrow \dfrac{-294-100}{9800}=\dfrac{1}{{{f}_{1}}}\left( \dfrac{98-25}{98} \right) \\ \Rightarrow \dfrac{-394}{100}=\dfrac{73}{{{f}_{1}}} \\ \Rightarrow \dfrac{1}{{{f}_{1}}}=\dfrac{-394}{7300}=-0.054\text{ m}{{\text{m}}^{-1}} $ And power is given as $P=\dfrac{-1}{{{f}_{1}}\left( \text{in m} \right)}=-\left( -0.054\times {{10}^{3}} \right) \\ \therefore P=5.4\text{ D}$