Question

question in attachment​

Answer 1

Power of lenses of spectacles of Kavita = -2.5 dioptre

a) Which lenses are used in her spectacles?

• => Diverging lens or concave lens or dispersive lens are used in her spectacles.

(b) State the defect of vision Kavita is suffering from.

• => Kavita is suffering from myopia (or Nearsightedness).

(c) Find the focal length of the lenses used in her spectacles.

We know that:-

• $\tt \longrightarrow \large f = \dfrac{1}{p} \\ \\ \tt where \: \: f = focal \: length \: and \: p = power$

• $\tt \rightarrow f = \dfrac{1}{ - 2.5}$
• $\tt \rightarrow f = \dfrac{1 0}{ - 25} = \dfrac{ - 2}{ 5}$
• => The focal length of the lenses used in her spectacles = -2/5

Myopic Eye :-

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Far point of myopic eye :-

$\setlength{\unitlength}{1.2cm}\begin{picture}(0,0)\thicklines\qbezier(0,1.72)(1.5,1.5)(1.67,0)\qbezier(1.67,0)(1.5,-1.5)(0,-1.72)\qbezier(0,1.72)(-1.2,1.7)(-1.7,1)\qbezier(0,-1.72)(-1.2,-1.7)(-1.7,-1)\qbezier(-1.7,1)(-2.3,0.7)(-2.3,0)\qbezier(-1.7,-1)(-2.3,-0.7)(-2.3,0)\qbezier(-1.7,1)(-2.13,0)(-1.7,-1)\qbezier(-1.7,1)(-1.15,0)(-1.7,-1)\qbezier(1.7,0)(1.7,0)(-5,0)\qbezier( - 5,0)( - 5,0)( -1.7,0.9)\qbezier(1.7,0)(1.7,0)( -1.7,0.9)\qbezier( - 5,0)( - 5,0)( -1.7, - 0.9)\qbezier(1.7,0)(1.7,0)( -1.7, - 0.9)\put(0.3,0.34){\vector(2, -1){0}}\put(0.3, -0.34){\vector( 2,1){0}}\put(0.34,0){\vector(2,0){0}}\put( -3.1,0.54){\vector(2, 1){0}}\put( - 3.1, -0.54){\vector(2, - 1){0}}\put( -3.1,0){\vector(2,0){0}}\put( -6.3, - 0.1){\sf{Far point}}\put(2, - 0.1){\sf{Retina}}\end{picture}$

Answer 2

Answer:

Power of lenses of spectacles of Kavita = -2.5 dioptre

a) Which lenses are used in her spectacles?

=> Diverging lens or concave lens or dispersive lens are used in her spectacles.

(b) State the defect of vision Kavita is suffering from.

=> Kavita is suffering from myopia (or Nearsightedness).

(c) Find the focal length of the lenses used in her spectacles.

We know that:-

$\tt \longrightarrow \large f = \dfrac{1}{p} \\ \\ \tt where \: \: f = focal \: length \: and \: p = power$

$\tt \rightarrow f = \dfrac{1}{ - 2.5}$

$\tt \rightarrow f = \dfrac{1 0}{ - 25} = \dfrac{ - 2}{ 5}$

=> The focal length of the lenses used in her spectacles = -2/5

Myopic Eye :-

$\setlength{\unitlength}{1.2cm}\begin{picture}(0,0)\thicklines\qbezier(0,1.72)(1.5,1.5)(1.67,0)\qbezier(1.67,0)(1.5,-1.5)(0,-1.72)\qbezier(0,1.72)(-1.2,1.7)(-1.7,1)\qbezier(0,-1.72)(-1.2,-1.7)(-1.7,-1)\qbezier(-1.7,1)(-2.3,0.7)(-2.3,0)\qbezier(-1.7,-1)(-2.3,-0.7)(-2.3,0)\qbezier(-1.7,1)(-2.13,0)(-1.7,-1)\qbezier(-1.7,1)(-1.15,0)(-1.7,-1)\put(-5,0){\line(1,0){5}}\put(-5,1.0){\line(1,0){3.3}}\put(-5,-1.0){\line(1,0){3.3}}\qbezier(-1.7,-1)(-1.7,-0.9)(0,0)\qbezier(-1.7,1)(-1.7,1)(0,0)\put(0,0){\line(1,0){1.7}}\put(-1,0){\vector(2,0){0}}\put(-1,0.6){\vector(2,-1){0}}\put(-1.1,-0.6){\vector(2,1){0}}\put(-3,0){\vector(1,0){0}}\put(-3,1){\vector(1,0){0}}\put(-3,-1){\vector(1,0){0}}\put(0,0){\circle{0.1}}\put(2,0){\large\bf Retina}\end{picture}$

Far point of myopic eye :-

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