Answer
415.2k+ views
Hint: Use lens formula to determine the image distance.
Differentiate the lens formula to determine the formula for rate of movement of image with respect to time.
Formula used:
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Here, f is the focal length, v is the image distance and u is the object distance.
Complete step by step answer:
We know that the focal length of the convex lens is positive. Also, the object distance from the lens is taken as negative.
We have the lens formula,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Here, f is the focal length, v is the image distance and u is the object distance.
Rearrange the above equation to determine the image distance as follows,
\[v = \dfrac{{fu}}{{u + f}}\]
Substitute 10 cm for f and \[ - 15\,cm\] for u in the above equation.
\[v = \dfrac{{\left( {10\,cm} \right)\left( { - 15\,cm} \right)}}{{ - 15\,cm + 10\,cm}}\]
Differentiate the lens formula with respect to time t as follows,
\[0 = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}\]
\[ \Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} = \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}\]
\[\therefore \dfrac{{dv}}{{dt}} = \dfrac{{{v^2}}}{{{u^2}}}\dfrac{{du}}{{dt}}\]
Substitute \[ + 30\,cm\] for v, \[ - 15\,cm\] for u and \[ + 2\,cm - {s^{ - 2}}\] for \[\dfrac{{du}}{{dt}}\] in the above equation.
\[\dfrac{{dv}}{{dt}} = \dfrac{{{{\left( {30\,cm} \right)}^2}}}{{{{\left( { - 15\,cm} \right)}^2}}}\left( {2\,cm - {s^{ - 2}}} \right)\]
\[\dfrac{{dv}}{{dt}} = \left( 4 \right)\left( {2\,cm - {s^{ - 2}}} \right)\]
\[\dfrac{{dv}}{{dt}} = + 8\,cm - {s^{ - 2}}\]
The positive sign in the above equation implies that the image is moving along the positive direction of the x-axis which is away from the lens
So, the correct answer is “Option D”.
Note:
Read the properties of the image formed by the convex lens. The focal length of the convex lens is positive whereas the focal length of the concave lens is negative. The object distance or source distance is negative while the image distance is positive.
Differentiate the lens formula to determine the formula for rate of movement of image with respect to time.
Formula used:
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Here, f is the focal length, v is the image distance and u is the object distance.
Complete step by step answer:
We know that the focal length of the convex lens is positive. Also, the object distance from the lens is taken as negative.
We have the lens formula,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Here, f is the focal length, v is the image distance and u is the object distance.
Rearrange the above equation to determine the image distance as follows,
\[v = \dfrac{{fu}}{{u + f}}\]
Substitute 10 cm for f and \[ - 15\,cm\] for u in the above equation.
\[v = \dfrac{{\left( {10\,cm} \right)\left( { - 15\,cm} \right)}}{{ - 15\,cm + 10\,cm}}\]
Differentiate the lens formula with respect to time t as follows,
\[0 = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}\]
\[ \Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} = \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}\]
\[\therefore \dfrac{{dv}}{{dt}} = \dfrac{{{v^2}}}{{{u^2}}}\dfrac{{du}}{{dt}}\]
Substitute \[ + 30\,cm\] for v, \[ - 15\,cm\] for u and \[ + 2\,cm - {s^{ - 2}}\] for \[\dfrac{{du}}{{dt}}\] in the above equation.
\[\dfrac{{dv}}{{dt}} = \dfrac{{{{\left( {30\,cm} \right)}^2}}}{{{{\left( { - 15\,cm} \right)}^2}}}\left( {2\,cm - {s^{ - 2}}} \right)\]
\[\dfrac{{dv}}{{dt}} = \left( 4 \right)\left( {2\,cm - {s^{ - 2}}} \right)\]
\[\dfrac{{dv}}{{dt}} = + 8\,cm - {s^{ - 2}}\]
The positive sign in the above equation implies that the image is moving along the positive direction of the x-axis which is away from the lens
So, the correct answer is “Option D”.
Note:
Read the properties of the image formed by the convex lens. The focal length of the convex lens is positive whereas the focal length of the concave lens is negative. The object distance or source distance is negative while the image distance is positive.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)