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1.An object place at 0.06 m from a convex lens of focal length 0.1 m. Calculate the position of the image?
पोकस रंफाई 0.1 भीटय के उत्तर रेंस से 0.06 भीटय ऩय एक वस्त ुयखी है । प्रततबफफं की स्स्थतत की गणना कयें?
![Page 3: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/3.jpg)
Ans. Here, u = - 0.06m = - 6 cm, f = 0.1 m = 10 cm, v =?
As 𝟏
𝒗 -
𝟏
𝒖 =
𝟏
𝒇
𝟏𝒗 =
𝟏
𝒇 +
𝟏
𝒖
= 𝟏
𝟏𝟎 -
𝟏
𝟔 =
𝟑−𝟓
𝟑𝟎 =
−𝟐
𝟑𝟎
= - 𝟏
𝟏𝟓 cm
v= - 15cm
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2) An object is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30cm. Find the position and nature of the image. एक वस्तु को 30 सेभी की बिज्मा के उत्तर दऩपण के साभने 20 सेभी की दयूी ऩय यखा गमा है। प्रततबफफं की स्स्थतत औय प्रकृतत का ऩता रगाएं।
![Page 5: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/5.jpg)
Ans. Here, object distance, u = - 20 cm, radius of curvature, R = 30 cm, image distance, v =?
As 𝟏
𝒗 +
𝟏
𝒖 =
𝟏
𝒇 =
𝟐
𝑹,
𝟏𝒗 =
𝟐
𝑹 -
𝟏
𝒖
𝟏𝒗 =
𝟐
𝟑𝟎 +
𝟏
𝟐𝟎 =
𝟒+𝟑
𝟔𝟎 =
𝟕
𝟔𝟎
v = 𝟔𝟎
𝟕 = 8.57
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As v is positive, image is at the back of the mirror. It must be virtual and erect.
ूं क v = सकाया भक है, प्रततबफफं दऩपण के ऩी े है।मह बासी औय
सीधा होना ा हए।
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3) Light enters from air into diamond, which has a refractive index of 2.42. Calculate the speed of light in diamond. The speed of light in air is 3 × 108 m/s.
प्रकाश हवा से हीये भें प्रवेश कयता है, स्िसका अऩवतपक सू कांक 2.42 है। हीये भें प्रकाश की गतत की गणना कयें। हवा भें प्रकाश की गतत 3 × 108 m/s है।
![Page 8: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/8.jpg)
Ans. Here, n = 2.42, v = ?, c= 3 × 108 m/s
Formula: n = 𝒄
𝒗 ,
v = 𝒄
𝒏 =
𝟑×𝟏𝟎𝟖
𝟐.𝟒𝟐 =1.24 × 10
8 m/s.
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4) With respect to air, the refractive index of ice is 1.31 and that of rock salt is 1.54. Calculate the refractive index of rock salt w.r.t. ice. हवा के साऩेऺ भें, फपप का अऩवतपनांक 1.31 है औय सेंधा नभक का अऩवतपनांक 1.54 है। यॉक साल्ट के अऩवतपनांक की गणना फपप के साऩेऺ कयें।
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Ans. Here, ni = 1.31 and nr = 1.54,
inr = ?
Now, inr
𝒂𝒏𝒓
𝒂𝒏𝒊 = 𝟏.𝟓𝟒
𝟏.𝟑𝟏 = 1.17
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5) A concave lens of focal length 40 cm is placed in contact with a concave lens of focal length 25 cm. What is the power the combination? पोकस रंफाई 40 सेभी के एक अवतर रेंस को पोकस रंफाई 25 सेभी के अवतर रेंस के संऩकप भें यखा िाता है। संमोिन के शस्तत की गणना कयें?
![Page 12: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/12.jpg)
Ans. Here f1 = 4 cm: f2 = - 25 cm; P =?
As P1= 𝟏𝟎𝟎
𝒇𝟏 =
𝟏𝟎𝟎
−𝟐𝟓 2.5 D
And P2= 𝟏𝟎𝟎
𝒇𝟐 =
𝟏𝟎𝟎
−𝟐𝟓 =- 4.0 D
As P = P1 +P2,
P = 2.5 – 4.0
= - 1.5 D
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6) A concave lens has focal length of 30 cm. Calculate at what distance should the object be placed from the lens so that it forms an image at 60 cm on the other side of the lens? Find the magnification produced by the lens in this case. एक अवतर रेंस की पोकस रंफाई 30 सेभी होती है। इस फात की गणना कयें क रेंस से वस्तु को कतनी दयूी ऩय यखा िाना ा हए ता क मह रेंस के दसूयी तयप 60 सेभी ऩय एक प्रततबफफं फना सके? रेंस द्वाया उ ऩा दत वधपन की गणना कयें?
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Ans. Here, F = 30 cm, u =? , v = 60 cm, m =?
As 𝟏
𝒇 =
𝟏
𝒗 -
𝟏
𝒖, ∴
𝟏
𝒖 =
𝟏
𝒗 -
𝟏
𝒇
= 𝟏
𝟔𝟎 -
𝟏
𝟑𝟎
= 𝟏−𝟐
𝟔𝟎 =
−𝟏
𝟔𝟎
u = - 60 cm
Thus, m=𝒉𝟐
𝒉𝟏 =
𝒖
𝒗 =
𝟔𝟎
−𝟔𝟎 = - 1
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7) An object is placed at a distance of 12 cm in front of a concave mirror. It forms a real image four times larger than the object. Calculate the distance of the image from the mirror.
कसी वस्तु को अवतर दऩपण के साभने 12 सेभी की दयूी ऩय यखा िाता है। मह वस्तु से ाय गुना फडी एक वास्तववक प्रततबफफं फनाता है। दऩपण से प्रततबफफं की दयूी की गणना कयें।
![Page 16: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/16.jpg)
Ans. Here, u = - 12 cm, m = - 4, v = ?
As m = −𝒗
𝒖 ;
-4 = −𝒗
−𝟏𝟐,
v= - 48 cm.
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8) A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also, find its magnification. एक 2.0 सेभी रंफी को पोकस रंफाई 10 सेभी के उत्तर रेंस के प्रभुख अऺ के रफंवत यखा गमा है। रेंस से वस्तु की दयूी 15 सेभी है। प्रततबफफं की प्रकृतत, स्स्थतत औय काय का ऩता रगाएं। इसके अरावा, इसका वधपन ऻात कीस्िए।
![Page 18: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/18.jpg)
Solution. Here, object size, h1 = 2.0 cm, focal length of convex lens,
f = 10 cm
Object distance, h1 = 2.0 cm, Image distance, v = ?, Image
size, h2 = ?
As 𝟏
𝒗 −
𝟏
𝒗 =
𝟏
𝒇,
𝟏
𝒗 =
𝟏
𝒇 +
𝟏
𝒖
= 𝟏
𝟏𝟎 −
𝟏
𝟏𝟓
= 𝟏
𝟑𝟎
v =30 cm.
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As v is positive, the image formed is on the right side of the lens. It must be real and inverted. िैसा क v सकाया भक है, ग ित प्रततबफफं रेंस के दाईं ओय है। मह वास्तववक औय उल्टा होना ा हए।
![Page 20: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/20.jpg)
As linear magnification m = 𝒉𝟐
𝒉𝟏 =
𝒗
𝒖,
𝒉𝟐
𝟐.𝟎 =
𝟑𝟎
−𝟏𝟓 = - 2
Or h2 = - 4.0 cm.
Negative sign of m and h2 show that the image is inverted.
M औय h2 के ऋणा भक ह से ऩता रता है क प्रततबफफं उरटी है।
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9) A concave lens has focal lenth of 15 cm at what distance should an object from the lens be placed so that it forms an image at 10 cm from the lens. find magnification? एक अवतर रेंस की पोकस दयूी 15 सेभी है, रेंस से कसी वस्तु को कतनी दयूी ऩय यखा िाना ा हए ता क वह रेंस से 10 सेभी की दयूी ऩय एक प्रततबफफं फनाए। वधपन का ऩता रगाएं
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Here, focal length of concave lens, f = - 15 cm.
Object distance, u =?, image distance, v = - 10 cm magnification
of lens, m=?
As 𝟏
𝒇 −
𝟏
𝒗 =
𝟏
𝒖,
𝟏
𝒖 =
𝟏
𝒗 +
𝟏
𝒇 =
𝟏
−𝟏𝟎 +
𝟏
𝟏𝟓 =
−𝟏
𝟑𝟎 or u = - 30 cm
Thus, the object should be placed at a distance of 30 cm on the left
side of the concave lens.
इस प्रकाय, वस्तु को अवतर रेंस के फाईं ओय 30 सेभी की दयूी ऩय यखा िाना ा हए।
![Page 23: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/23.jpg)
Linear magnification, m = 𝒖
𝒗 =
−𝟏𝟎
−𝟑𝟎 =
𝟏
𝟑 .
The positive sign of m shows that the image is virtual and erect, and
its size is (1/3) of the size of the object.
m का सकाया भक भान दशापता है क प्रततबफफं बासी औय सीधा है, औय इसका काय वस्तु के काय का (1/3) है।
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10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of this combination? Also, calculate focal length of the combination. पोकस रफंाई 25 सेभी का एक अवतर रेंस औय पोकस रफंाई 20 सेभी का उत्तर रेंस एक दसूये के संऩकप भें त ेहैं। इस संमोिन की शस्तत तमा है? इसके अरावा, संमोिन की पोकस रफंाई की गणना कयें।
![Page 25: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/25.jpg)
Solution. Here, focal length of concave lens, f1 = - 25 cm, focal
length of convex lens, f2 = + cm
power of the combination, P = ?,
Focal length of the combination, F = ?
Positive sign of P and F indicates that the combination of two given
lenses behaves as a convex lens.
P औय F का सकाया भक संकेत इं गत कयता है क दो दए गए रेंस का संमोिन उत्तर रेंस के ऩ भें मवहाय कयता है।
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11) A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 10 cm. What if the focal length and power of the combination?
पोकस रंफाई 20 सेभी के उत्तर रेंस को पोकस रंफाई 10 सेभी के अवतर रेंस के संऩकप भें यखा िाता है। संमोिन की पोकस रंफाई औय शस्तत तमा होगा?
![Page 27: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/27.jpg)
Solution. Here, f1 = + 20 cm, f2 = - 10 cm; F = ?, P = ?
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12) A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. एक 4.5 सेभी सईु को पोकर रफंाई 15 सेभी के उत्तर दऩपण से 12 सेभी दयू यखा गमा है। प्रततबफफं औय वधपन की स्स्थतत ऻात कयें।
![Page 29: An object place at 0.06 m from a convex...10) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed contact with each other. What is the power of](https://reader033.vdocuments.us/reader033/viewer/2022061513/5e74e7c20586ab01ba5abe37/html5/thumbnails/29.jpg)
Solution. Here, object size, h1 = 4.5 cm Object distance, u = - 12 cm Focal length, f = + 15 cmImage distance, v = ? (to be
calculated)Magnification, m = ?
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13) An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed. एक 2.5 सेभी ऊं ा तीय, 25 सेभी की दयूी ऩय 20 सेभी पोकर रंफाई के अऩसायी दऩपण के साभने यखा गमा है। फनाई गई प्रततबफफं की प्रकृतत, स्स्थतत औय काय का ऩता रगाएं।
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Solution. Here, Object size, h1 = + 2.5 cm,object distance, u = - 25 cm Focal length of diverging mirror, f = + 20 cm Image distance, v = ?, image size h2=?
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14) Find the focal length of a convex mirror of radius of curvature 1 m?वक्रता बिज्मा 1 भीटय के उत्तर दऩपण की पोकर रंफाई ऻात कयें?
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Solution. Here, focal length, f =?
Radius of curvature, R = 1 m (+ for convex
mirror)
As f = R/2.
∴ f = 1/2 m = 0.5 m
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15) The image formed by a convex mirror of focal length 20 cm is a quarter of the object. What is the distance of the object from the mirror?पोकर रंफाई 20 सेभी के उत्तर दऩपण द्वाया फना प्रततबफफं वस्त ुका एक ौथाई है। दऩपण से वस्त ुकी दयूी तमा है?
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16) A real image, 1/5th of size of the object is formed at a distance of 18 cm from a mirror. What is the nature of mirror? Calculate its focal length.
एक वास्तववक प्रततबफफं, वस्तु का काय का 1/5th दऩपण से 18 सेभी की दयूी ऩय फनता है। दऩपण की प्रकृतत तमा है? इसकी पोकर रंफाई की गणना कयें।
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Image is real. It is concave mirror. प्रततबफफं वास्तववक है। मह अवतर दऩपण है।
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17) An object 4 cm high if placed at a distance of 6 cm in from of a concave mirror of focal length 12 cm. Find the position of image.एक 4 सेभी ऊं ी वस्त ुम द पोकर रफंाई 12 सेभी के अवतर दऩपण से 6 सेभी की दयूी ऩय यखी िाती है। प्रततबफफं की स्स्थतत ऻात कीस्िए?
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v=4 The image will be formed between pole and focus.
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18)A compound lens is made up of two thin lenses giving power + 12.5 D and – 2.5D.Find the focal length and power of the combination.एक संमोिी रेंस दो ऩतरे रेंस से फना होता है ,स्िनकी ऺभता + 12.5 D औय - 2.5D देता है। संमोिन की पोकर रंफाई औय ऺभता ऻात कयें।
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19) Light enters from air to kerosene having a refractive index of 1.47. what is speed of light in kerosene.
एक प्रकाश कयण हवा से केयोससन भें प्रवेश कयता है स्िसभें 1.47 का अऩवतपनांक होता है। सभट्टी के तेर भें प्रकाश की गतत तमा है।
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20) A 5 cm tall object is placed perpendicular to principal axis of a convex lens of focal length 10 cm. If the object is place 30 cm away from the lens, find position, size and nature of image. एक 5 सेभी रंफी वस्तु 10 सेभी पोकस रफंाई के उत्तर रेंस के प्रभुख अऺ के रंफवत यखा गमा है। म द वस्तु रेंस से 30 सेभी की दयूी ऩय है, तो प्रततबफफं की स्स्थतत, काय औय प्रकृतत की गणना कयें।
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21.find the focal length of the lens of
power -2.0D .What type of lens is
this?-2.0 D ऩॉवय के रेंस के पोकस दयूी की गणना कयें। मह कस प्रकाय का रेंस है?
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𝐟𝐨𝐫𝐦𝐮𝐥𝐚 − 𝐟𝐨𝐜𝐚𝐥 𝐥𝐞𝐧𝐠𝐭𝐡= 𝟏
𝐩𝐨𝐰𝐞𝐫
f=− 𝟏
𝟐
f= -0.5m=-50cm
since focal length is negative ,it is a concave lens.
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22.A doctor has prescribeed a corrective lens of power +1.5D. Find the focal lenth of the lens, is the prescribed a conversing and diverging lens?एक डॉतटय ने + 1.5D ऩावय का सधुाया भक रेंस तनधापरयत कमा है। रेंस का पोकस दयूी ऻात कीस्िए, असबसायी है मा अऩसायी रेंस है?
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Formula- 𝐟𝐨𝐜𝐚𝐥 𝐥𝐞𝐧𝐭𝐡= 𝟏
𝐩𝐨𝐰𝐞𝐫
f= 𝟏
𝟏.𝟓=0.666m=66.6cm
focal lenth is positive it is converging lens.
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23. Speed of light in water is?ऩानी भे प्रकाश की ार होती है?
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Formula- speed in a medium = 𝐬𝐩𝐞𝐞𝐝 𝐨𝐟 𝐥𝐢𝐠𝐡𝐭
𝐫𝐞𝐟𝐥𝐞𝐜𝐭𝐢𝐯𝐞 𝐢𝐧𝐝𝐞𝐱 𝐨𝐟 𝐭𝐡𝐞 𝐦𝐞𝐝𝐢𝐮𝐦
Reflective index of the water is =4/3
speed of light in water/ ऩानी भे प्रकाश की गतत =
𝟒/𝟑
2.26×108 m/s
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24.Radius of curvature of a lens is 20cm,find the focal length of the lens? एक रेंस की वक्रता का बिज्मा 20सेभी है, रेंस का पोकस दयूी ऻात कयें?
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FORMULA - Radius of curvature = 2 × focal lenth
Focal lenth = 𝑹
𝟐
= 𝟐𝟎
𝟐 = 10cm
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25. Find the Radius of curvature if the focal length of lens is 15cm?
रेंस की वक्रता का बिज्मा ऻात कयें म द रेंस की पोकस दयूी 15 सेभी है?
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FORMULA- Radius of curvature = 2×focal lenth
= 2×15
= 30cm
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THANK YOU