Physics, asked by AliceJoy, 1 year ago

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Question for the known ones only

An eye specialist suggested a +2D lens to the person with defect in the vision .
Which kind of defect in the vision does he have ?

Draw diagrams to show the defect of vision and its correction with suitable lens.

Answers

Answered by Anjula
219

\boxed{ANSWER} :-

•He has an eye defect named \bold{Hypermetropia}.

•As the value of power of lens is positive .

\boxed{Hypermetropia}

—> It is called as Far sightedness

—> A person with this defect can’t see object at near distances clearly.

—> The focal length of eye lens formed is greater than 2.27cm

—> When rays comes from near objects ,after refraction at eye lens ,the image is formed beyond retina.

—> For people of this defect,the eye lens can form the image on retina when object is placed beyond far point.

—>So to correct this defect we use the eye lens which forms an image of an object beyond near point.The object is between near point and least distance of distinct vision.

—> We use biconcave lens to correct defect of hypermetropia

Refer attachment :)

Attachments:
Answered by Anonymous
253

 \huge{ \frak {\red{ \underline {\underline{Your \: Answer :-}}}}}

 \rightarrow \mathtt{ \: If \: the \: person \: uses \: the \: lens \: of }

  \tt{power +2 \: Dioptre (positive).}

 \rightarrow \tt{Then \: the \: person  \: is  \: using \:  the  \: Convex \:  lens.}

 \rightarrow \tt{The \:  convex  \: lens \:  is  \: used  \: to \:  correct  }

 \tt{the  \: Hypermetropia  \: (Far -sightedness).}

Hypermetropia is the defect of vision in which a person sees faraway objects clearly but cannot see the nearby objects.

 \frak{The  \: image  \: is  \:  formed \:  behind \: }\frak{retina \:  in  \:  \bold{Hypermetropia.}}

 \boxed{\large{ \green{ \star\: Causes : }}}

1 ] The distance between the eye lens and retina decreases on account of either shortening of eyeball or flattening of lens.

2 ] Weak action of ciliary muscles causes low converging power of lens.

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