Step 1: Finding the image formed by the first lens . For lens A, the object distance is \( u_1 = -20.0 \, \text{cm} \) (since the object is real) and the focal length is \( f_1 = 10.0 \, \text{cm} \). Using the lens formula: \[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1} \] \[ \frac{1}{10.0} = \frac{1}{v_1} - \frac{1}{-20.0} \] \[ v_1 = 20.0 \, \text{cm} \] So, the image formed by lens A is at \( v_1 = 20.0 \, \text{cm} \), which is real and inverted.
Step 2: Finding the image formed by the second lens . The image formed by lens A acts as the object for lens B. The object distance for lens B is the distance between the two lenses, i.e., \( u_2 = 70.0 - 20.0 = 50.0 \, \text{cm} \). Using the lens formula for lens B: \[ \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \] \[ \frac{1}{10.0} = \frac{1}{v_2} - \frac{1}{50.0} \] \[ v_2 = 12.5 \, \text{cm} \] Thus, the final image is formed at a distance of 12.5 cm from lens B, which is real and inverted.
A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(i)} Express the distance \( y \) between the wall and foot of the ladder in terms of \( h \) and height \( x \) on the wall at a certain instant. Also, write an expression in terms of \( h \) and \( x \) for the area \( A \) of the right triangle, as seen from the side by an observer.
निम्नलिखित गद्यांश की सप्रसंग व्याख्या कीजिए :
‘‘पुर्ज़े खोलकर फिर ठीक करना उतना कठिन काम नहीं है, लोग सीखते भी हैं, सिखाते भी हैं, अनाड़ी के हाथ में चाहे घड़ी मत दो पर जो घड़ीसाज़ी का इम्तहान पास कर आया है उसे तो देखने दो । साथ ही यह भी समझा दो कि आपको स्वयं घड़ी देखना, साफ़ करना और सुधारना आता है कि नहीं । हमें तो धोखा होता है कि परदादा की घड़ी जेब में डाले फिरते हो, वह बंद हो गई है, तुम्हें न चाबी देना आता है न पुर्ज़े सुधारना तो भी दूसरों को हाथ नहीं लगाने देते इत्यादि ।’’