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    make a diagram to show how hypermetropia is corrected. the near point of a hypermetropic eye is 1 m. what is the power of the lens required to correct this defect? assume that the near point of the normal eye is 25 cm.

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    get make a diagram to show how hypermetropia is corrected. the near point of a hypermetropic eye is 1 m. what is the power of the lens required to correct this defect? assume that the near point of the normal eye is 25 cm. from screen.

    Make a diagram to show how hypermetropia is corrected. The near point of hypermetropic eye is 1 m. What is the power of lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

    Click hereπŸ‘†to get an answer to your question ✍️ Make a diagram to show how hypermetropia is corrected. The near point of hypermetropic eye is 1 m. What is the power of lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

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    Make a diagram to show how hypermetropia is corrected. The near point of hypermetropic eye is 1 m. What is the power of lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

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    Updated on : 2022-09-05

    Solution Verified by Toppr

    Hypermetropia can be corrected by using a convex lens. A convex lens converges the incoming light such that the image is formed on the retina.

    An object at 25 cm forms an image at the near point of the hypermetropic eye. Here, near point is 1 m.

    Given,

    Object distance,u=βˆ’25 cm

    Image distance, v=βˆ’100 cm

    From lens formula, v 1 ​ βˆ’ u 1 ​ = f 1 ​ βˆ’100 1 ​ βˆ’ βˆ’25 1 ​ = f 1 ​

    Focal length,f=100/3 cm=1/3 m

    Power, P= f 1 ​ = 1/3 1 ​ =3 D

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    Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

    Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm. - Get the answer to this question and access a vast question bank that is tailored for students.

    Academic QuestionsPhysics QuestionsMake A Diagram To Show How Hypermetropia Is Corrected The Near Point Of A Hypermetropic Eye Is 1 M What Is The Power Of The Lens Required To Correct This Defect

    Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

    Hypermetropia is also known as hyperopia or long-sightedness or far-sightedness. Hypermetropia is the condition of the eyes where the image of a nearby object is formed behind the retina. Here, the light is focused behind the retina instead of focusing on the retina. Hypermetropia is mainly caused due to certain structural defects in the retina. Structural defects comprise of:

    Small-sized eye-ball

    Non-circular lenses

    The cornea is flatter than usual

    Defective blood vessels in the retina

    The convex lens creates a virtual image of a nearby object (N’ in the above figure) at the near point of vision (N) of the individual suffering from hypermetropia.

    The given individual will be able to clearly see the object kept at 25 cm (near the point of the normal eye) if the image of the object is formed at his near point, which is given as 1 m.

    Object distance, u= – 25 cm

    Image distance, v= – 1 m = – 100 m

    Focal length, f

    Using the lens formula,

    1vβˆ’1u=1f = βˆ’1100βˆ’1βˆ’25=1f = 1f=βˆ’125βˆ’1100 =

    1f=βˆ’4βˆ’1100=3100=0.33m

    Power= 1/f = 1/0.33 = +3.0 D

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    Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 metre. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm.

    A person suffering from hypermetropia can see distinct objecdts clearly but faces difficulty is seeing nearly objects clearly. It happens because the eye lense focuses the incoming diverget rays beyond the ratina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following firgure . The convex lens actually creates a virtual image of a nearby object (N'in hte figure ) at the near point of vision (N) of the perosn suffering from hyermetropia. The given perons wil be able to clearly see the object kept at 25 cm (near point of the normal eye). if the image of the object is formed at his near point, which is given as 1 m Object distance , u=-25 cm Image distance , v=-1m =-100m Focal length ,f Using the lens formula, (1)/(v)-(1)/(u)=(1)/(f) -(1)/(100)-(1)/(-25)=(1)/(f) (1)/(f)=-(1)/(25)=(1)/(100) (1)/(f)=-(4-1)/(100) f=(100)/(3)=33.3 cm=0.33m We known, Powr , P=(1)/(f(" in meteres")) P=(1)/(0.33)=+30.0D A convex lens of power +3.0 D is required to correct the defect.

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    Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is

    1metre 1metre

    . What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is

    25cm 25cm .

    Updated On: 27-06-2022

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    Text Solution Open Answer in App Solution

    A person suffering from hypermetropia can see distinct objecdts clearly but faces difficulty is seeing nearly objects clearly. It happens because the eye lense focuses the incoming diverget rays beyond the ratina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following firgure .

    The convex lens actually creates a virtual image of a nearby object (N'in hte figure ) at the near point of vision (N) of the perosn suffering from hyermetropia.

    The given perons wil be able to clearly see the object kept at 25 cm (near point of the normal eye). if the image of the object is formed at his near point, which is given as 1 m

    Object distance , u=-25 cm

    Image distance , v=-1m =-100m

    Focal length ,f

    Using the lens formula,

    1 v βˆ’ 1 u = 1 f 1v-1u=1f βˆ’ 1 100 βˆ’ 1 βˆ’25 = 1 f -1100-1-25=1f 1 f =βˆ’ 1 25 = 1 100 1f=-125=1100 1 f =βˆ’ 4βˆ’1 100 1f=-4-1100 f= 100 3 =33.3cm=0.33m f=1003=33.3cm=0.33m We known, Powr , P= 1 f( in meteres) P=1f( in meteres) P= 1 0.33 =+30.0D P=10.33=+30.0D

    A convex lens of power +3.0 D is required to correct the defect.

    Answer

    Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

    Transcript

    hello friends in the person we have to make a diagram to show hypermetropia is corrected ok the near point of a hypermetropic eye is given here as one metre and where to find the power of lens required to correct this defect ok and we have to assume that in your point of the normalised 25cm so in hypermetropia ok in hypermetropia the images formed behind the retina letters in image formed on the retina behind ok behind retina object of its focus at point P which is behind the retina ok so we have to increase the converging power of the islands ok and to do so we introduce

    dance in front of the islands ok let's say that we introduce the convex lens like this ok in front of the US as a country for this configuration the image of an object placed at O you can see the image of an object placed at home that is the normal near point of the eye is formed on retina and for the man it it appears to be at at the point of the man ok object is at the near point of normalised that is 25 CM OK then the image formed is at order that the near point of the men and hear the year

    1 metre is equal to we will be -103 needs to find formula we know that 1 upon 3 minus 1 upon 2 is equal to 1 upon from here we can say one upon this because this that lets you search CV 200 CM upon my hundred minus one upon us -25 and this will be equal to 1 upon focal length of a convex lens used ok that we can say this minus 5 minus plus sandal get one upon 25 - 100 to 1 upon

    focal length f OK so the power will be What power will be one upon f and one upon us is as a hundred ok and those that the current 4 - 1 + 1 upon F3 upon hundred universe ok to convert it into m inverse will get World War I will multiply it with with hundred and that is will get three major inverse X hundred ok and does will get power as one upon us and power will be able to diopter it and see that the connections will be used with the power of three drive axle for this question guys thank you

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